3.68 \(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=119 \[ \frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {7 x}{8 a^3}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

-7/8*x/a^3-I*ln(cos(d*x+c))/a^3/d-1/6*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^3+3/8*I*tan(d*x+c)^2/a/d/(a+I*a*tan(d*
x+c))^2+7/8*I/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.24, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3558, 3595, 3589, 3475, 12, 3526, 8} \[ \frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {7 x}{8 a^3}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-7*x)/(8*a^3) - (I*Log[Cos[c + d*x]])/(a^3*d) - Tan[c + d*x]^3/(6*d*(a + I*a*Tan[c + d*x])^3) + (((3*I)/8)*Ta
n[c + d*x]^2)/(a*d*(a + I*a*Tan[c + d*x])^2) + ((7*I)/8)/(d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3475

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

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c -
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rule 3595

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^2(c+d x) (-3 a+6 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan (c+d x) \left (-18 i a^2-24 a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {i \int \frac {42 a^3 \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{24 a^5}+\frac {i \int \tan (c+d x) \, dx}{a^3}\\ &=-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {(7 i) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {7 \int 1 \, dx}{8 a^3}\\ &=-\frac {7 x}{8 a^3}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 118, normalized size = 0.99 \[ \frac {\sec ^3(c+d x) (-81 i \sin (c+d x)+84 d x \sin (3 (c+d x))+2 i \sin (3 (c+d x))-51 \cos (c+d x)+\cos (3 (c+d x)) (96 \log (\cos (c+d x))-84 i d x-2)+96 i \sin (3 (c+d x)) \log (\cos (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(-51*Cos[c + d*x] + Cos[3*(c + d*x)]*(-2 - (84*I)*d*x + 96*Log[Cos[c + d*x]]) - (81*I)*Sin[c +
d*x] + (2*I)*Sin[3*(c + d*x)] + 84*d*x*Sin[3*(c + d*x)] + (96*I)*Log[Cos[c + d*x]]*Sin[3*(c + d*x)]))/(96*a^3
*d*(-I + Tan[c + d*x])^3)

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fricas [A]  time = 0.42, size = 77, normalized size = 0.65 \[ -\frac {{\left (180 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 i \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 66 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(180*d*x*e^(6*I*d*x + 6*I*c) + 96*I*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 66*I*e^(4*I*d*x +
4*I*c) + 15*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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giac [A]  time = 4.09, size = 80, normalized size = 0.67 \[ -\frac {-\frac {90 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {165 i \, \tan \left (d x + c\right )^{3} + 291 \, \tan \left (d x + c\right )^{2} - 171 i \, \tan \left (d x + c\right ) - 29}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(-90*I*log(tan(d*x + c) - I)/a^3 - 6*I*log(-I*tan(d*x + c) + 1)/a^3 + (165*I*tan(d*x + c)^3 + 291*tan(d*
x + c)^2 - 171*I*tan(d*x + c) - 29)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [A]  time = 0.15, size = 98, normalized size = 0.82 \[ \frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right )}{16 a^{3} d}-\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17}{8 a^{3} d \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x)

[Out]

1/16*I/d/a^3*ln(tan(d*x+c)+I)+7/8*I/d/a^3/(tan(d*x+c)-I)^2+15/16*I/d/a^3*ln(tan(d*x+c)-I)-1/6/d/a^3/(tan(d*x+c
)-I)^3+17/8/d/a^3/(tan(d*x+c)-I)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 3.85, size = 110, normalized size = 0.92 \[ -\frac {\frac {27\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^3}-\frac {17{}\mathrm {i}}{12\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,17{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,15{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(log(tan(c + d*x) - 1i)*15i)/(16*a^3*d) - ((27*tan(c + d*x))/(8*a^3) - 17i/(12*a^3) + (tan(c + d*x)^2*17i)/(8*
a^3))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (log(tan(c + d*x) + 1i)*1i)/(16*a^3*d
)

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sympy [A]  time = 0.58, size = 187, normalized size = 1.57 \[ \begin {cases} - \frac {\left (- 16896 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 3840 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 15 e^{6 i c} + 11 e^{4 i c} - 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {15}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {15 x}{8 a^{3}} - \frac {i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise((-(-16896*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 3840*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 512*I*
a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*((
-15*exp(6*I*c) + 11*exp(4*I*c) - 5*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) + 15/(8*a**3)), True)) - 15*x/(8*a**3)
- I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)

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