\(\int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\) [99]

Optimal result

Integrand size = 28, antiderivative size = 205 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {5 a b^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {2 b^5 \cos (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {b^5 \sec (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 205, 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, 2713, 2645, 30, 2644, 2672, 308, 212, 2670, 276} \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {a^5 \sin ^3(c+d x)}{3 d}+\frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos ^3(c+d x)}{3 d}+\frac {10 a^3 b^2 \sin ^3(c+d x)}{3 d}+\frac {10 a^2 b^3 \cos ^3(c+d x)}{3 d}-\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {5 a b^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {5 a b^4 \sin ^3(c+d x)}{3 d}-\frac {5 a b^4 \sin (c+d x)}{d}-\frac {b^5 \cos ^3(c+d x)}{3 d}+\frac {2 b^5 \cos (c+d x)}{d}+\frac {b^5 \sec (c+d x)}{d} \]

[In]

Int[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

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 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(632\) vs. \(2(205)=410\).

Time = 8.36 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.08 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {b^5 \cos ^5(c+d x) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4+30 a^2 b^2-7 b^4\right ) \cos ^6(c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos ^5(c+d x) \cos (3 (c+d x)) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {5 a b^4 \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {5 a b^4 \cos ^5(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^5}{d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}-\frac {b^5 \cos ^5(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^5}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (3 a^4+10 a^2 b^2-25 b^4\right ) \cos ^5(c+d x) \sin (c+d x) (a+b \tan (c+d x))^5}{4 d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac {a \left (a^4-10 a^2 b^2+5 b^4\right ) \cos ^5(c+d x) \sin (3 (c+d x)) (a+b \tan (c+d x))^5}{12 d (a \cos (c+d x)+b \sin (c+d x))^5} \]

[In]

Integrate[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(b^5*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 + 30*a^2*b^2 -
7*b^4)*Cos[c + d*x]^6*(a + b*Tan[c + d*x])^5)/(4*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]^5*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5)
- (5*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)/(d*(a*Cos[c + d*x]
+ b*Sin[c + d*x])^5) + (5*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
)/(d*(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)/(d*(C
os[(c + d*x)/2] - Sin[(c + d*x)/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)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) + (a*(
3*a^4 + 10*a^2*b^2 - 25*b^4)*Cos[c + d*x]^5*Sin[c + d*x]*(a + b*Tan[c + d*x])^5)/(4*d*(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[3*(c + d*x)]*(a + b*Tan[c + d*x])^5)/(12*d*(a*
Cos[c + d*x] + b*Sin[c + d*x])^5)

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {5 a^{4} b \cos \left (d x +c \right )^{3}}{3}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {10 a^{2} b^{3} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+5 a \,b^{4} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(169\)
default	\(\frac {\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}-\frac {5 a^{4} b \cos \left (d x +c \right )^{3}}{3}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {10 a^{2} b^{3} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+5 a \,b^{4} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(169\)
parts	\(\frac {a^{5} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {b^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3 d}+\frac {5 a \,b^{4} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {10 a^{2} b^{3} \left (\frac {\cos \left (d x +c \right )^{3}}{3}-\cos \left (d x +c \right )\right )}{d}-\frac {5 a^{4} b}{3 \sec \left (d x +c \right )^{3} d}\)	\(186\)
parallelrisch	\(\frac {-120 a \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+120 a \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+20 \left (-a^{4} b -4 a^{2} b^{3}+b^{5}\right ) \cos \left (2 d x +2 c \right )+\left (-5 a^{4} b +10 a^{2} b^{3}-b^{5}\right ) \cos \left (4 d x +4 c \right )+10 \left (a^{5}+2 a^{3} b^{2}-7 a \,b^{4}\right ) \sin \left (2 d x +2 c \right )+a \left (a^{4}-10 a^{2} b^{2}+5 b^{4}\right ) \sin \left (4 d x +4 c \right )-40 b \left (\left (a^{4}+4 a^{2} b^{2}-\frac {8}{5} b^{4}\right ) \cos \left (d x +c \right )+\frac {3 a^{4}}{8}+\frac {9 a^{2} b^{2}}{4}-\frac {9 b^{4}}{8}\right )}{24 d \cos \left (d x +c \right )}\)	\(225\)
risch	\(-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )} a^{4} b}{8 d}-\frac {15 \,{\mathrm e}^{i \left (d x +c \right )} a^{2} b^{3}}{4 d}+\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b^{5}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{5}}{8 d}-\frac {25 i {\mathrm e}^{-i \left (d x +c \right )} a \,b^{4}}{8 d}+\frac {25 i {\mathrm e}^{i \left (d x +c \right )} a \,b^{4}}{8 d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{4} b}{8 d}-\frac {15 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2} b^{3}}{4 d}+\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b^{5}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{5}}{8 d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{3} b^{2}}{4 d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} b^{2}}{4 d}+\frac {2 b^{5} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {5 a \,b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {5 a \,b^{4} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {5 b \cos \left (3 d x +3 c \right ) a^{4}}{12 d}+\frac {5 b^{3} \cos \left (3 d x +3 c \right ) a^{2}}{6 d}-\frac {b^{5} \cos \left (3 d x +3 c \right )}{12 d}+\frac {a^{5} \sin \left (3 d x +3 c \right )}{12 d}-\frac {5 a^{3} \sin \left (3 d x +3 c \right ) b^{2}}{6 d}+\frac {5 a \sin \left (3 d x +3 c \right ) b^{4}}{12 d}\)	\(412\)
norman	\(\frac {\frac {10 a^{4} b +40 a^{2} b^{3}-16 b^{5}}{3 d}-\frac {10 a^{4} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {5 \left (2 a^{4} b +8 a^{2} b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {5 \left (4 a^{4} b +16 a^{2} b^{3}-16 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {2 \left (5 a^{4} b +80 a^{2} b^{3}-32 b^{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {2 a \left (a^{4}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a \left (a^{4}-5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {10 a \left (a^{4}+8 a^{2} b^{2}-13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {10 a \left (a^{4}+8 a^{2} b^{2}-13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {4 a \left (a^{4}+20 a^{2} b^{2}-25 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {4 a \left (a^{4}+20 a^{2} b^{2}-25 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 b \left (5 a^{4}-40 a^{2} b^{2}-8 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {5 a \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {5 a \,b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(446\)

[In]

int(sec(d*x+c)^2*(cos(d*x+c)*a+b*sin(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/3*a^5*(2+cos(d*x+c)^2)*sin(d*x+c)-5/3*a^4*b*cos(d*x+c)^3+10/3*a^3*b^2*sin(d*x+c)^3-10/3*a^2*b^3*(2+sin(
d*x+c)^2)*cos(d*x+c)+5*a*b^4*(-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+b^5*(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 = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, a b^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a b^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, b^{5} - 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{5} + 5 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**2*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.79 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=-\frac {10 \, a^{4} b \cos \left (d x + c\right )^{3} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5} - 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b^{3} + 5 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b^{4} + 2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{5}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/6*(10*a^4*b*cos(d*x + c)^3 - 20*a^3*b^2*sin(d*x + c)^3 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^5 - 20*(cos(
d*x + c)^3 - 3*cos(d*x + c))*a^2*b^3 + 5*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)
+ 6*sin(d*x + c))*a*b^4 + 2*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*b^5)/d

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.38 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, b^{5}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 50 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b - 20 \, a^{2} b^{3} + 5 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/3*(15*a*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*b^5/(tan(1/
2*d*x + 1/2*c)^2 - 1) + 2*(3*a^5*tan(1/2*d*x + 1/2*c)^5 - 15*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^4*b*tan(1/2*d
*x + 1/2*c)^4 + 3*b^5*tan(1/2*d*x + 1/2*c)^4 + 2*a^5*tan(1/2*d*x + 1/2*c)^3 + 40*a^3*b^2*tan(1/2*d*x + 1/2*c)^
3 - 50*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 + 12*b^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^
5*tan(1/2*d*x + 1/2*c) - 15*a*b^4*tan(1/2*d*x + 1/2*c) - 5*a^4*b - 20*a^2*b^3 + 5*b^5)/(tan(1/2*d*x + 1/2*c)^2
+ 1)^3)/d

Mupad [B] (verification not implemented)

Time = 26.78 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.35 \[ \int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {10\,a\,b^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (10\,a\,b^4-2\,a^5\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^4\,b-40\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {2\,a^5}{3}-\frac {80\,a^3\,b^2}{3}+\frac {70\,a\,b^4}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {10\,a^4\,b}{3}-\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+\frac {10\,a^4\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (10\,a\,b^4-2\,a^5\right )-\frac {16\,b^5}{3}+\frac {40\,a^2\,b^3}{3}-10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^2,x)

[Out]

(10*a*b^4*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)*(10*a*b^4 - 2*a^5) + tan(c/2 + (d*x)/2)^4*(10*a^4
*b - 40*a^2*b^3) + tan(c/2 + (d*x)/2)^3*((70*a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^5*((7
0*a*b^4)/3 + (2*a^5)/3 - (80*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^2*((10*a^4*b)/3 + (32*b^5)/3 - (80*a^2*b^3)/3) +
(10*a^4*b)/3 - tan(c/2 + (d*x)/2)^7*(10*a*b^4 - 2*a^5) - (16*b^5)/3 + (40*a^2*b^3)/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))