\(\int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\) [101]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {10 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {15 a b^4 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 b \cos (c+d x)}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {b^5 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}-\frac {2 b^5 \sec (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d} \]

[Out]

10*a^3*b^2*arctanh(sin(d*x+c))/d-15/2*a*b^4*arctanh(sin(d*x+c))/d-5*a^4*b*cos(d*x+c)/d+10*a^2*b^3*cos(d*x+c)/d
-b^5*cos(d*x+c)/d+10*a^2*b^3*sec(d*x+c)/d-2*b^5*sec(d*x+c)/d+1/3*b^5*sec(d*x+c)^3/d+a^5*sin(d*x+c)/d-10*a^3*b^
2*sin(d*x+c)/d+15/2*a*b^4*sin(d*x+c)/d+5/2*a*b^4*sin(d*x+c)*tan(d*x+c)^2/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3169, 2717, 2718, 2672, 327, 212, 2670, 14, 294, 276} \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos (c+d x)}{d}+\frac {10 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}-\frac {15 a b^4 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {b^5 \cos (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}-\frac {2 b^5 \sec (c+d x)}{d} \]

[In]

Int[Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(10*a^3*b^2*ArcTanh[Sin[c + d*x]])/d - (15*a*b^4*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*b*Cos[c + d*x])/d + (10
*a^2*b^3*Cos[c + d*x])/d - (b^5*Cos[c + d*x])/d + (10*a^2*b^3*Sec[c + d*x])/d - (2*b^5*Sec[c + d*x])/d + (b^5*
Sec[c + d*x]^3)/(3*d) + (a^5*Sin[c + d*x])/d - (10*a^3*b^2*Sin[c + d*x])/d + (15*a*b^4*Sin[c + d*x])/(2*d) + (
5*a*b^4*Sin[c + d*x]*Tan[c + d*x]^2)/(2*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3169

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^5 \cos (c+d x)+5 a^4 b \sin (c+d x)+10 a^3 b^2 \sin (c+d x) \tan (c+d x)+10 a^2 b^3 \sin (c+d x) \tan ^2(c+d x)+5 a b^4 \sin (c+d x) \tan ^3(c+d x)+b^5 \sin (c+d x) \tan ^4(c+d x)\right ) \, dx \\ & = a^5 \int \cos (c+d x) \, dx+\left (5 a^4 b\right ) \int \sin (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sin (c+d x) \tan (c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sin (c+d x) \tan ^2(c+d x) \, dx+\left (5 a b^4\right ) \int \sin (c+d x) \tan ^3(c+d x) \, dx+b^5 \int \sin (c+d x) \tan ^4(c+d x) \, dx \\ & = -\frac {5 a^4 b \cos (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}+\frac {\left (10 a^3 b^2\right ) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a^2 b^3\right ) \text {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (5 a b^4\right ) \text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^5 \text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {5 a^4 b \cos (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}+\frac {\left (10 a^3 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a^2 b^3\right ) \text {Subst}\left (\int \left (-1+\frac {1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (15 a b^4\right ) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}-\frac {b^5 \text {Subst}\left (\int \left (1+\frac {1}{x^4}-\frac {2}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {10 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a^4 b \cos (c+d x)}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {b^5 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}-\frac {2 b^5 \sec (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {\left (15 a b^4\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d} \\ & = \frac {10 a^3 b^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {15 a b^4 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 b \cos (c+d x)}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}-\frac {b^5 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}-\frac {2 b^5 \sec (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(892\) vs. \(2(204)=408\).

Time = 7.69 (sec) , antiderivative size = 892, normalized size of antiderivative = 4.37 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {b^3 \left (-60 a^2+11 b^2\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{6 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {5 \left (4 a^3 b^2-3 a b^4\right ) \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{2 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {5 \left (4 a^3 b^2-3 a b^4\right ) \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{2 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {\left (15 a b^4+b^5\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {\left (-15 a b^4+b^5\right ) \cos ^5(c+d x) (a+b \tan (c+d x))^5}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {\cos ^5(c+d x) \left (60 a^2 b^3 \sin \left (\frac {1}{2} (c+d x)\right )-11 b^5 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {\cos ^5(c+d x) \left (-60 a^2 b^3 \sin \left (\frac {1}{2} (c+d x)\right )+11 b^5 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (a^4-10 a^2 b^2+5 b^4\right ) \cos ^5(c+d x) \sin (c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5} \]

[In]

Integrate[Sec[c + d*x]^4*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

-1/6*(b^3*(-60*a^2 + 11*b^2)*Cos[c + d*x]^5*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) -
(b*(5*a^4 - 10*a^2*b^2 + b^4)*Cos[c + d*x]^6*(a + b*Tan[c + d*x])^5)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) -
(5*(4*a^3*b^2 - 3*a*b^4)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^5)/(2*d
*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (5*(4*a^3*b^2 - 3*a*b^4)*Cos[c + d*x]^5*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]]*(a + b*Tan[c + d*x])^5)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((15*a*b^4 + b^5)*Cos[c + d*x]^
5*(a + b*Tan[c + d*x])^5)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) +
(b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(a*
Cos[c + d*x] + b*Sin[c + d*x])^5) - (b^5*Cos[c + d*x]^5*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^5)/(6*d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + ((-15*a*b^4 + b^5)*Cos[c + d*x]^5*(a +
b*Tan[c + d*x])^5)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c
+ d*x]^5*(60*a^2*b^3*Sin[(c + d*x)/2] - 11*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(6*d*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (Cos[c + d*x]^5*(-60*a^2*b^3*Sin[(c + d*x)/2] +
11*b^5*Sin[(c + d*x)/2])*(a + b*Tan[c + d*x])^5)/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] +
b*Sin[c + d*x])^5) + (a*(a^4 - 10*a^2*b^2 + 5*b^4)*Cos[c + d*x]^5*Sin[c + d*x]*(a + b*Tan[c + d*x])^5)/(d*(a*C
os[c + d*x] + b*Sin[c + d*x])^5)

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {a^{5} \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) a^{4} b +10 a^{3} b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+10 a^{2} b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+5 a \,b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(230\)
default	\(\frac {a^{5} \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) a^{4} b +10 a^{3} b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+10 a^{2} b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+5 a \,b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(230\)
parts	\(\frac {a^{5} \sin \left (d x +c \right )}{d}+\frac {b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {10 a^{3} b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {5 a \,b^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {10 a^{2} b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}-\frac {5 a^{4} b \cos \left (d x +c \right )}{d}\)	\(244\)
parallelrisch	\(\frac {-180 b^{2} \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a \left (a^{2}-\frac {3 b^{2}}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+180 b^{2} \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) a \left (a^{2}-\frac {3 b^{2}}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (-15 a^{4} b +60 a^{2} b^{3}-8 b^{5}\right ) \cos \left (3 d x +3 c \right )+12 \left (-5 a^{4} b +20 a^{2} b^{3}-3 b^{5}\right ) \cos \left (2 d x +2 c \right )+3 \left (-5 a^{4} b +10 a^{2} b^{3}-b^{5}\right ) \cos \left (4 d x +4 c \right )+6 \left (a^{5}-10 a^{3} b^{2}+10 a \,b^{4}\right ) \sin \left (2 d x +2 c \right )+3 \left (a^{5}-10 a^{3} b^{2}+5 a \,b^{4}\right ) \sin \left (4 d x +4 c \right )-90 b \left (\left (a^{4}-4 a^{2} b^{2}+\frac {8}{15} b^{4}\right ) \cos \left (d x +c \right )+\frac {a^{4}}{2}-\frac {7 a^{2} b^{2}}{3}+\frac {5 b^{4}}{18}\right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(314\)
risch	\(-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{4} b}{2 d}+\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b^{3}}{d}-\frac {{\mathrm e}^{i \left (d x +c \right )} b^{5}}{2 d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a \,b^{4}}{2 d}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b^{2}}{d}+\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b^{2}}{d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{4} b}{2 d}+\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{3}}{d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{5}}{2 d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a \,b^{4}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{5}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{5}}{2 d}-\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )} \left (15 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-60 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+16 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-15 i a b -60 a^{2}+12 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {10 a^{3} b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {15 a \,b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {10 a^{3} b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {15 a \,b^{4} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}\)	\(460\)
norman	\(\frac {\frac {a \left (2 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}+\frac {a \left (2 a^{4}-20 a^{2} b^{2}+35 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {a \left (6 a^{4}-60 a^{2} b^{2}+5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {30 a^{4} b -120 a^{2} b^{3}+16 b^{5}}{3 d}-\frac {10 a^{4} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}-\frac {2 \left (5 a^{4} b +20 a^{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {2 \left (15 a^{4} b -120 a^{2} b^{3}+16 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {2 \left (45 a^{4} b -240 a^{2} b^{3}+64 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {2 \left (45 a^{4} b -120 a^{2} b^{3}-16 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}-\frac {2 \left (45 a^{4} b -60 a^{2} b^{3}+16 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {3 a \left (2 a^{4}-20 a^{2} b^{2}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {3 a \left (2 a^{4}-20 a^{2} b^{2}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {a \left (2 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (2 a^{4}-20 a^{2} b^{2}+35 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a \left (6 a^{4}-60 a^{2} b^{2}+5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 b \left (45 a^{4}+60 a^{2} b^{2}-56 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {5 a \,b^{2} \left (4 a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {5 a \,b^{2} \left (4 a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(637\)

[In]

int(sec(d*x+c)^4*(cos(d*x+c)*a+b*sin(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^5*sin(d*x+c)-5*cos(d*x+c)*a^4*b+10*a^3*b^2*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+10*a^2*b^3*(sin(d*x+
c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+5*a*b^4*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d
*x+c)-3/2*ln(sec(d*x+c)+tan(d*x+c)))+b^5*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c
)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.93 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {4 \, b^{5} - 12 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 24 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (5 \, a b^{4} \cos \left (d x + c\right ) + 2 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/12*(4*b^5 - 12*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 15*(4*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3*log(sin
(d*x + c) + 1) - 15*(4*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 24*(5*a^2*b^3 - b^5)*cos(d*x
+ c)^2 + 6*(5*a*b^4*cos(d*x + c) + 2*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x +
c)^3)

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.89 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {15 \, a b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} - 120 \, a^{2} b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 4 \, b^{5} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - 60 \, a^{3} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 60 \, a^{4} b \cos \left (d x + c\right ) - 12 \, a^{5} \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/12*(15*a*b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*s
in(d*x + c)) - 120*a^2*b^3*(1/cos(d*x + c) + cos(d*x + c)) + 4*b^5*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*
cos(d*x + c)) - 60*a^3*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)) + 60*a^4*b*cos(d*x
+ c) - 12*a^5*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.38 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {12 \, {\left (a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b + 10 \, a^{2} b^{3} - b^{5}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, a^{2} b^{3} + 10 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/6*(15*(4*a^3*b^2 - 3*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3*b^2 - 3*a*b^4)*log(abs(tan(1/2*d*
x + 1/2*c) - 1)) + 12*(a^5*tan(1/2*d*x + 1/2*c) - 10*a^3*b^2*tan(1/2*d*x + 1/2*c) + 5*a*b^4*tan(1/2*d*x + 1/2*
c) - 5*a^4*b + 10*a^2*b^3 - b^5)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(15*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 60*a^2*b^
3*tan(1/2*d*x + 1/2*c)^4 + 6*b^5*tan(1/2*d*x + 1/2*c)^4 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 24*b^5*tan(1/2*
d*x + 1/2*c)^2 - 15*a*b^4*tan(1/2*d*x + 1/2*c) - 60*a^2*b^3 + 10*b^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 26.85 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (15\,a\,b^4-20\,a^3\,b^2\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (30\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^5-60\,a^3\,b^2+25\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^5-60\,a^3\,b^2+25\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (30\,a^4\,b-80\,a^2\,b^3+\frac {32\,b^5}{3}\right )-10\,a^4\,b-\frac {16\,b^5}{3}+40\,a^2\,b^3+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^4,x)

[Out]

- (atanh(tan(c/2 + (d*x)/2))*(15*a*b^4 - 20*a^3*b^2))/d - (tan(c/2 + (d*x)/2)*(15*a*b^4 + 2*a^5 - 20*a^3*b^2)
- tan(c/2 + (d*x)/2)^4*(30*a^4*b - 40*a^2*b^3) - tan(c/2 + (d*x)/2)^7*(15*a*b^4 + 2*a^5 - 20*a^3*b^2) - tan(c/
2 + (d*x)/2)^3*(25*a*b^4 + 6*a^5 - 60*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(25*a*b^4 + 6*a^5 - 60*a^3*b^2) + tan(c/
2 + (d*x)/2)^2*(30*a^4*b + (32*b^5)/3 - 80*a^2*b^3) - 10*a^4*b - (16*b^5)/3 + 40*a^2*b^3 + 10*a^4*b*tan(c/2 +
(d*x)/2)^6)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 - 1))