3.1.63 \(\int (a+i a \tan (c+d x))^5 \, dx\) [63]

Optimal. Leaf size=117 \[ 16 a^5 x-\frac {16 i a^5 \log (\cos (c+d x))}{d}-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\frac {2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]

[Out]

16*a^5*x-16*I*a^5*ln(cos(d*x+c))/d-8*a^5*tan(d*x+c)/d+2/3*I*a^2*(a+I*a*tan(d*x+c))^3/d+1/4*I*a*(a+I*a*tan(d*x+
c))^4/d+2*I*a*(a^2+I*a^2*tan(d*x+c))^2/d

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Rubi [A]
time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3559, 3558, 3556} \begin {gather*} -\frac {8 a^5 \tan (c+d x)}{d}-\frac {16 i a^5 \log (\cos (c+d x))}{d}+16 a^5 x+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^5,x]

[Out]

16*a^5*x - ((16*I)*a^5*Log[Cos[c + d*x]])/d - (8*a^5*Tan[c + d*x])/d + (((2*I)/3)*a^2*(a + I*a*Tan[c + d*x])^3
)/d + ((I/4)*a*(a + I*a*Tan[c + d*x])^4)/d + ((2*I)*a*(a^2 + I*a^2*Tan[c + d*x])^2)/d

Rule 3556

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

Rule 3558

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[b^2*(Tan[c + d*x]/d), x]) /; FreeQ[{a, b, c, d}, x]

Rule 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rubi steps

\begin {align*} \int (a+i a \tan (c+d x))^5 \, dx &=\frac {i a (a+i a \tan (c+d x))^4}{4 d}+(2 a) \int (a+i a \tan (c+d x))^4 \, dx\\ &=\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (4 a^2\right ) \int (a+i a \tan (c+d x))^3 \, dx\\ &=\frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (8 a^3\right ) \int (a+i a \tan (c+d x))^2 \, dx\\ &=16 a^5 x-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (16 i a^5\right ) \int \tan (c+d x) \, dx\\ &=16 a^5 x-\frac {16 i a^5 \log (\cos (c+d x))}{d}-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}\\ \end {align*}

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Mathematica [A]
time = 2.50, size = 228, normalized size = 1.95 \begin {gather*} \frac {a^5 \sec (c) \sec ^4(c+d x) \left (-18 i \cos (3 c+2 d x)+48 d x \cos (3 c+2 d x)+12 d x \cos (3 c+4 d x)+12 d x \cos (5 c+4 d x)+6 \cos (c+2 d x) \left (-3 i+8 d x-4 i \log \left (\cos ^2(c+d x)\right )\right )+\cos (c) \left (-33 i+72 d x-36 i \log \left (\cos ^2(c+d x)\right )\right )-24 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )-6 i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-6 i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+75 \sin (c)-70 \sin (c+2 d x)+30 \sin (3 c+2 d x)-25 \sin (3 c+4 d x)\right )}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*Sec[c]*Sec[c + d*x]^4*((-18*I)*Cos[3*c + 2*d*x] + 48*d*x*Cos[3*c + 2*d*x] + 12*d*x*Cos[3*c + 4*d*x] + 12*
d*x*Cos[5*c + 4*d*x] + 6*Cos[c + 2*d*x]*(-3*I + 8*d*x - (4*I)*Log[Cos[c + d*x]^2]) + Cos[c]*(-33*I + 72*d*x -
(36*I)*Log[Cos[c + d*x]^2]) - (24*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (6*I)*Cos[3*c + 4*d*x]*Log[Cos[c +
d*x]^2] - (6*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 75*Sin[c] - 70*Sin[c + 2*d*x] + 30*Sin[3*c + 2*d*x] -
25*Sin[3*c + 4*d*x]))/(12*d)

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Maple [A]
time = 0.05, size = 72, normalized size = 0.62

method	result	size
derivativedivides	\(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {11 i \left (\tan ^{2}\left (d x +c \right )\right )}{2}+8 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(72\)
default	\(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {11 i \left (\tan ^{2}\left (d x +c \right )\right )}{2}+8 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(72\)
risch	\(-\frac {32 a^{5} c}{d}-\frac {4 i a^{5} \left (48 \,{\mathrm e}^{6 i \left (d x +c \right )}+108 \,{\mathrm e}^{4 i \left (d x +c \right )}+88 \,{\mathrm e}^{2 i \left (d x +c \right )}+25\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {16 i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(89\)
norman	\(16 a^{5} x -\frac {15 a^{5} \tan \left (d x +c \right )}{d}+\frac {5 a^{5} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {11 i a^{5} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {i a^{5} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {8 i a^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/d*a^5*(-15*tan(d*x+c)+1/4*I*tan(d*x+c)^4+5/3*tan(d*x+c)^3-11/2*I*tan(d*x+c)^2+8*I*ln(1+tan(d*x+c)^2)+16*arct
an(tan(d*x+c)))

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Maxima [A]
time = 0.51, size = 165, normalized size = 1.41 \begin {gather*} a^{5} x + \frac {5 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{5}}{3 \, d} + \frac {10 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{5}}{d} + \frac {i \, a^{5} {\left (\frac {4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{4 \, d} + \frac {5 i \, a^{5} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {5 i \, a^{5} \log \left (\sec \left (d x + c\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

a^5*x + 5/3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^5/d + 10*(d*x + c - tan(d*x + c))*a^5/d + 1/4*I*
a^5*((4*sin(d*x + c)^2 - 3)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 2*log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*(
1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1))/d + 5*I*a^5*log(sec(d*x + c))/d

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Fricas [A]
time = 0.36, size = 177, normalized size = 1.51 \begin {gather*} -\frac {4 \, {\left (48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 108 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 25 i \, a^{5} + 12 \, {\left (i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

-4/3*(48*I*a^5*e^(6*I*d*x + 6*I*c) + 108*I*a^5*e^(4*I*d*x + 4*I*c) + 88*I*a^5*e^(2*I*d*x + 2*I*c) + 25*I*a^5 +
12*(I*a^5*e^(8*I*d*x + 8*I*c) + 4*I*a^5*e^(6*I*d*x + 6*I*c) + 6*I*a^5*e^(4*I*d*x + 4*I*c) + 4*I*a^5*e^(2*I*d*
x + 2*I*c) + I*a^5)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*
I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]
time = 0.26, size = 178, normalized size = 1.52 \begin {gather*} - \frac {16 i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 192 i a^{5} e^{6 i c} e^{6 i d x} - 432 i a^{5} e^{4 i c} e^{4 i d x} - 352 i a^{5} e^{2 i c} e^{2 i d x} - 100 i a^{5}}{3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} + 3 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**5,x)

[Out]

-16*I*a**5*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-192*I*a**5*exp(6*I*c)*exp(6*I*d*x) - 432*I*a**5*exp(4*I*c)*ex
p(4*I*d*x) - 352*I*a**5*exp(2*I*c)*exp(2*I*d*x) - 100*I*a**5)/(3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*e
xp(6*I*d*x) + 18*d*exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (99) = 198\).
time = 0.56, size = 222, normalized size = 1.90 \begin {gather*} -\frac {4 \, {\left (12 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 72 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 108 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 25 i \, a^{5}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

-4/3*(12*I*a^5*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 48*I*a^5*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 72*I*a^5*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 48*I*a^5*e^(2*I*d*x + 2*I*c)*log(e
^(2*I*d*x + 2*I*c) + 1) + 48*I*a^5*e^(6*I*d*x + 6*I*c) + 108*I*a^5*e^(4*I*d*x + 4*I*c) + 88*I*a^5*e^(2*I*d*x +
2*I*c) + 12*I*a^5*log(e^(2*I*d*x + 2*I*c) + 1) + 25*I*a^5)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) +
6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

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Mupad [B]
time = 3.27, size = 73, normalized size = 0.62 \begin {gather*} \frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}-15\,a^5\,\mathrm {tan}\left (c+d\,x\right )-\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\,11{}\mathrm {i}}{2}+\frac {5\,a^5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^5,x)

[Out]

(a^5*log(tan(c + d*x) + 1i)*16i - 15*a^5*tan(c + d*x) - (a^5*tan(c + d*x)^2*11i)/2 + (5*a^5*tan(c + d*x)^3)/3
+ (a^5*tan(c + d*x)^4*1i)/4)/d

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