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

Optimal. Leaf size=204 \[ \frac {a^5 \sin (c+d x)}{d}-\frac {5 a^4 b \cos (c+d x)}{d}-\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {10 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^5 \cos (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}-\frac {2 b^5 \sec (c+d x)}{d} \]

[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

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Rubi [A]  time = 0.21, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3090, 2637, 2638, 2592, 321, 206, 2590, 14, 288, 270} \[ -\frac {10 a^3 b^2 \sin (c+d x)}{d}+\frac {10 a^2 b^3 \cos (c+d x)}{d}+\frac {10 a^2 b^3 \sec (c+d x)}{d}+\frac {10 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 a^4 b \cos (c+d x)}{d}+\frac {a^5 \sin (c+d x)}{d}+\frac {15 a b^4 \sin (c+d x)}{2 d}+\frac {5 a b^4 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b^5 \cos (c+d x)}{d}+\frac {b^5 \sec ^3(c+d x)}{3 d}-\frac {2 b^5 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[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 206

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

Rule 270

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 288

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 321

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 2590

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 2592

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 2637

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

Rule 2638

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

Rule 3090

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*} \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\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 ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (5 a b^4\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^5 \operatorname {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 ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a^2 b^3\right ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}-\frac {b^5 \operatorname {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 \tanh ^{-1}(\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 ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac {10 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {15 a b^4 \tanh ^{-1}(\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*}

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Mathematica [A]  time = 5.90, size = 397, normalized size = 1.95 \[ \frac {120 a^2 b^3-30 a b^2 \left (4 a^2-3 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 a b^2 \left (4 a^2-3 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 b^3 \left (60 a^2-11 b^2\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 b^3 \left (11 b^2-60 a^2\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+12 a \left (a^4-10 a^2 b^2+5 b^4\right ) \sin (c+d x)-12 b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos (c+d x)+\frac {b^4 (15 a+b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^4 (b-15 a)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^5 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {2 b^5 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}-22 b^5}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a^2*b^3 - 22*b^5 - 12*b*(5*a^4 - 10*a^2*b^2 + b^4)*Cos[c + d*x] - 30*a*b^2*(4*a^2 - 3*b^2)*Log[Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]] + 30*a*b^2*(4*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^4*(15*a +
b))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^5*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
+ (2*b^3*(60*a^2 - 11*b^2)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^5*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (b^4*(-15*a + b))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*b^3*
(-60*a^2 + 11*b^2)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*a*(a^4 - 10*a^2*b^2 + 5*b^4)*S
in[c + d*x])/(12*d)

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fricas [A]  time = 0.50, size = 190, normalized size = 0.93 \[ \frac {4 \, b^{5} - 12 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 24 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (5 \, a b^{4} \cos \left (d x + c\right ) + 2 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.60, size = 281, normalized size = 1.38 \[ \frac {15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {12 \, {\left (a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{4} b + 10 \, a^{2} b^{3} - b^{5}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, a^{2} b^{3} + 10 \, b^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.23, size = 327, normalized size = 1.60 \[ \frac {a^{5} \sin \left (d x +c \right )}{d}-\frac {5 a^{4} b \cos \left (d x +c \right )}{d}-\frac {10 a^{3} b^{2} \sin \left (d x +c \right )}{d}+\frac {10 a^{3} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {10 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {10 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{d}+\frac {20 a^{2} b^{3} \cos \left (d x +c \right )}{d}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d}+\frac {15 a \,b^{4} \sin \left (d x +c \right )}{2 d}-\frac {15 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}-\frac {8 b^{5} \cos \left (d x +c \right )}{3 d}-\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}-\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^5*sin(d*x+c)/d-5*a^4*b*cos(d*x+c)/d-10*a^3*b^2*sin(d*x+c)/d+10/d*a^3*b^2*ln(sec(d*x+c)+tan(d*x+c))+10/d*a^2*
b^3*sin(d*x+c)^4/cos(d*x+c)+10/d*cos(d*x+c)*sin(d*x+c)^2*a^2*b^3+20*a^2*b^3*cos(d*x+c)/d+5/2/d*a*b^4*sin(d*x+c
)^5/cos(d*x+c)^2+5/2*a*b^4*sin(d*x+c)^3/d+15/2*a*b^4*sin(d*x+c)/d-15/2/d*a*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/3/d
*b^5*sin(d*x+c)^6/cos(d*x+c)^3-1/d*b^5*sin(d*x+c)^6/cos(d*x+c)-8/3*b^5*cos(d*x+c)/d-1/d*b^5*cos(d*x+c)*sin(d*x
+c)^4-4/3/d*cos(d*x+c)*sin(d*x+c)^2*b^5

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maxima [A]  time = 0.33, size = 181, normalized size = 0.89 \[ -\frac {15 \, a b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} - 120 \, a^{2} b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 4 \, b^{5} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - 60 \, a^{3} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 60 \, a^{4} b \cos \left (d x + c\right ) - 12 \, a^{5} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.05, size = 302, normalized size = 1.48 \[ -\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (15\,a\,b^4-20\,a^3\,b^2\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (30\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (2\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^5-60\,a^3\,b^2+25\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^5-60\,a^3\,b^2+25\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (30\,a^4\,b-80\,a^2\,b^3+\frac {32\,b^5}{3}\right )-10\,a^4\,b-\frac {16\,b^5}{3}+40\,a^2\,b^3+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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