3.72 \(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx\)
Optimal. Leaf size=98 \[ \frac{5 i a^5 \sec (c+d x)}{d}+\frac{5 a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \]
[Out]
(5*a^5*ArcTanh[Sin[c + d*x]])/d + ((5*I)*a^5*Sec[c + d*x])/d + (((10*I)/3)*a^3*Cos[c + d*x]*(a + I*a*Tan[c + d
*x])^2)/d - (((2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4)/d
________________________________________________________________________________________
Rubi [A] time = 0.0892792, antiderivative size = 98, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used =
{3496, 3486, 3770} \[ \frac{5 i a^5 \sec (c+d x)}{d}+\frac{5 a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^5,x]
[Out]
(5*a^5*ArcTanh[Sin[c + d*x]])/d + ((5*I)*a^5*Sec[c + d*x])/d + (((10*I)/3)*a^3*Cos[c + d*x]*(a + I*a*Tan[c + d
*x])^2)/d - (((2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4)/d
Rule 3496
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Rule 3486
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \cos ^3(c+d x) (a+i a \tan (c+d x))^5 \, dx &=-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}-\frac{1}{3} \left (5 a^2\right ) \int \cos (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}+\left (5 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{5 i a^5 \sec (c+d x)}{d}+\frac{10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}+\left (5 a^5\right ) \int \sec (c+d x) \, dx\\ &=\frac{5 a^5 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 i a^5 \sec (c+d x)}{d}+\frac{10 i a^3 \cos (c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^4}{3 d}\\ \end{align*}
Mathematica [A] time = 1.63698, size = 130, normalized size = 1.33 \[ \frac{a^5 \cos ^4(c+d x) (\tan (c+d x)-i)^5 \left (30 (\sin (5 c)+i \cos (5 c)) \cos (c+d x) \tanh ^{-1}\left (\cos (c) \tan \left (\frac{d x}{2}\right )+\sin (c)\right )-(\cos (3 c-2 d x)-i \sin (3 c-2 d x)) (-17 i \sin (2 (c+d x))+13 \cos (2 (c+d x))+10)\right )}{3 d (\cos (d x)+i \sin (d x))^5} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^5,x]
[Out]
(a^5*Cos[c + d*x]^4*(30*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*Cos[c + d*x]*(I*Cos[5*c] + Sin[5*c]) - (Cos[3*c
- 2*d*x] - I*Sin[3*c - 2*d*x])*(10 + 13*Cos[2*(c + d*x)] - (17*I)*Sin[2*(c + d*x)]))*(-I + Tan[c + d*x])^5)/(3
*d*(Cos[d*x] + I*Sin[d*x])^5)
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Maple [A] time = 0.062, size = 179, normalized size = 1.8 \begin{align*}{\frac{i{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+{\frac{{\frac{28\,i}{3}}{a}^{5}\cos \left ( dx+c \right ) }{d}}+{\frac{i{a}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{{\frac{14\,i}{3}}{a}^{5}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-5\,{\frac{{a}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{13\,{a}^{5}\sin \left ( dx+c \right ) }{3\,d}}+5\,{\frac{{a}^{5}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{\frac{5\,i}{3}}{a}^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{5}}{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x)
[Out]
I/d*a^5*sin(d*x+c)^6/cos(d*x+c)+28/3*I/d*a^5*cos(d*x+c)+I/d*a^5*cos(d*x+c)*sin(d*x+c)^4+14/3*I/d*a^5*cos(d*x+c
)*sin(d*x+c)^2-5*a^5*sin(d*x+c)^3/d-13/3*a^5*sin(d*x+c)/d+5/d*a^5*ln(sec(d*x+c)+tan(d*x+c))-5/3*I/d*a^5*cos(d*
x+c)^3+1/3/d*sin(d*x+c)*cos(d*x+c)^2*a^5
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Maxima [A] time = 1.13882, size = 208, normalized size = 2.12 \begin{align*} -\frac{10 i \, a^{5} \cos \left (d x + c\right )^{3} + 20 \, a^{5} \sin \left (d x + c\right )^{3} + 2 i \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{5} + 20 i \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{5} + 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^{5} + 2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{5}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")
[Out]
-1/6*(10*I*a^5*cos(d*x + c)^3 + 20*a^5*sin(d*x + c)^3 + 2*I*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))
*a^5 + 20*I*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^5 + 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^5 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^5)/d
________________________________________________________________________________________
Fricas [A] time = 1.28215, size = 332, normalized size = 3.39 \begin{align*} \frac{-4 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 20 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 15 \,{\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \,{\left (a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")
[Out]
1/3*(-4*I*a^5*e^(5*I*d*x + 5*I*c) + 20*I*a^5*e^(3*I*d*x + 3*I*c) + 30*I*a^5*e^(I*d*x + I*c) + 15*(a^5*e^(2*I*d
*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) + I) - 15*(a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) - I))/(d*
e^(2*I*d*x + 2*I*c) + d)
________________________________________________________________________________________
Sympy [A] time = 0.977797, size = 150, normalized size = 1.53 \begin{align*} \frac{5 a^{5} \left (- \log{\left (e^{i d x} - i e^{- i c} \right )} + \log{\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \frac{2 i a^{5} e^{- i c} e^{i d x}}{d \left (e^{2 i d x} + e^{- 2 i c}\right )} + \begin{cases} \frac{- 4 i a^{5} d e^{3 i c} e^{3 i d x} + 24 i a^{5} d e^{i c} e^{i d x}}{3 d^{2}} & \text{for}\: 3 d^{2} \neq 0 \\x \left (4 a^{5} e^{3 i c} - 8 a^{5} e^{i c}\right ) & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**5,x)
[Out]
5*a**5*(-log(exp(I*d*x) - I*exp(-I*c)) + log(exp(I*d*x) + I*exp(-I*c)))/d + 2*I*a**5*exp(-I*c)*exp(I*d*x)/(d*(
exp(2*I*d*x) + exp(-2*I*c))) + Piecewise(((-4*I*a**5*d*exp(3*I*c)*exp(3*I*d*x) + 24*I*a**5*d*exp(I*c)*exp(I*d*
x))/(3*d**2), Ne(3*d**2, 0)), (x*(4*a**5*exp(3*I*c) - 8*a**5*exp(I*c)), True))
________________________________________________________________________________________
Giac [B] time = 2.05691, size = 2272, normalized size = 23.18 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")
[Out]
-1/6144*(39225*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 313800*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(
I*d*x + I*c) + 1) + 1098300*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(10*I*d*x + 2*
I*c)*log(I*e^(I*d*x + I*c) + 1) + 2196600*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1098300*a^5*e^(
4*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 313800*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2745
750*a^5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 39225*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 8520*a^5*e^
(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 68160*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 23
8560*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 477120*a^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*
c) - 1) + 477120*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 238560*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(
I*d*x + I*c) - 1) + 68160*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 596400*a^5*e^(8*I*d*x)*log(I*e^
(I*d*x + I*c) - 1) + 8520*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) - 1) - 39225*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^
(I*d*x + I*c) + 1) - 313800*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1098300*a^5*e^(12*I*d*x + 4
*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2196600*a^5
*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1098300*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1)
- 313800*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2745750*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c)
+ 1) - 39225*a^5*e^(-8*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 8520*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I*c
) - 1) - 68160*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 238560*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e
^(I*d*x + I*c) - 1) - 477120*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 477120*a^5*e^(6*I*d*x - 2*
I*c)*log(-I*e^(I*d*x + I*c) - 1) - 238560*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 68160*a^5*e^(2
*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 596400*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 8520*a^5*e^
(-8*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 15*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 120*a^5*e^(14
*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 420*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 840*a
^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 840*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c))
+ 420*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 120*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) + 1050*a^5*e^(8*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 15*a^5*e^(-8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 15*
a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 120*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x) + e^(-I*
c)) - 420*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 840*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x
) + e^(-I*c)) - 840*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 420*a^5*e^(4*I*d*x - 4*I*c)*log(-I*
e^(I*d*x) + e^(-I*c)) - 120*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 1050*a^5*e^(8*I*d*x)*log(-I
*e^(I*d*x) + e^(-I*c)) - 15*a^5*e^(-8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 8192*I*a^5*e^(19*I*d*x + 11*I*c) + 1
6384*I*a^5*e^(17*I*d*x + 9*I*c) - 176128*I*a^5*e^(15*I*d*x + 7*I*c) - 1003520*I*a^5*e^(13*I*d*x + 5*I*c) - 243
7120*I*a^5*e^(11*I*d*x + 3*I*c) - 3411968*I*a^5*e^(9*I*d*x + I*c) - 2953216*I*a^5*e^(7*I*d*x - I*c) - 1568768*
I*a^5*e^(5*I*d*x - 3*I*c) - 471040*I*a^5*e^(3*I*d*x - 5*I*c) - 61440*I*a^5*e^(I*d*x - 7*I*c))/(d*e^(16*I*d*x +
8*I*c) + 8*d*e^(14*I*d*x + 6*I*c) + 28*d*e^(12*I*d*x + 4*I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*I*d*x -
2*I*c) + 28*d*e^(4*I*d*x - 4*I*c) + 8*d*e^(2*I*d*x - 6*I*c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))