3.1.78 \(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [78]

Optimal. Leaf size=126 \[ \frac {i x}{16 a^4}+\frac {3}{16 a^4 d (1+i \tan (c+d x))}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2} \]

[Out]

1/16*I*x/a^4+3/16/a^4/d/(1+I*tan(d*x+c))+1/8*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^4+1/12*I*tan(d*x+c)^3/a/d/(a+I*
a*tan(d*x+c))^3-1/16/d/(a^2+I*a^2*tan(d*x+c))^2

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3628, 3627, 3621, 3607, 8} \begin {gather*} \frac {3}{16 a^4 d (1+i \tan (c+d x))}+\frac {i x}{16 a^4}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/16)*x)/a^4 + 3/(16*a^4*d*(1 + I*Tan[c + d*x])) + Tan[c + d*x]^4/(8*d*(a + I*a*Tan[c + d*x])^4) + ((I/12)*T
an[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^3) - 1/(16*d*(a^2 + I*a^2*Tan[c + d*x])^2)

Rule 8

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

Rule 3607

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 3621

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^m/(2*a^3*f*m)), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*
Simp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a
*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rule 3627

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*b*f*m)), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3628

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

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{2 a}\\ &=\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2}\\ &=\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {i \int \frac {a-2 i a \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{8 a^4}\\ &=\frac {3}{16 a^4 d (1+i \tan (c+d x))}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i \int 1 \, dx}{16 a^4}\\ &=\frac {i x}{16 a^4}+\frac {3}{16 a^4 d (1+i \tan (c+d x))}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.49, size = 95, normalized size = 0.75 \begin {gather*} \frac {\sec ^4(c+d x) (16 \cos (2 (c+d x))+3 (1+8 i d x) \cos (4 (c+d x))+32 i \sin (2 (c+d x))-3 i \sin (4 (c+d x))-24 d x \sin (4 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*(16*Cos[2*(c + d*x)] + 3*(1 + (8*I)*d*x)*Cos[4*(c + d*x)] + (32*I)*Sin[2*(c + d*x)] - (3*I)*Si
n[4*(c + d*x)] - 24*d*x*Sin[4*(c + d*x)]))/(384*a^4*d*(-I + Tan[c + d*x])^4)

________________________________________________________________________________________

Maple [A]
time = 0.15, size = 87, normalized size = 0.69

method	result	size
risch	\(\frac {i x}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{4} d}+\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\)	\(60\)
derivativedivides	\(\frac {\frac {i}{16 \tan \left (d x +c \right )-16 i}-\frac {5 i}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {7}{16 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right )}{32}-\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{32}}{d \,a^{4}}\)	\(87\)
default	\(\frac {\frac {i}{16 \tan \left (d x +c \right )-16 i}-\frac {5 i}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {7}{16 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right )}{32}-\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{32}}{d \,a^{4}}\)	\(87\)
norman	\(\frac {\frac {1}{12 d a}+\frac {i x}{16 a}-\frac {\tan ^{6}\left (d x +c \right )}{2 d a}+\frac {3 \left (\tan ^{4}\left (d x +c \right )\right )}{4 d a}+\frac {\tan ^{2}\left (d x +c \right )}{3 d a}+\frac {i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {i x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}-\frac {i \tan \left (d x +c \right )}{16 d a}-\frac {11 i \left (\tan ^{3}\left (d x +c \right )\right )}{48 d a}-\frac {53 i \left (\tan ^{5}\left (d x +c \right )\right )}{48 d a}+\frac {i \left (\tan ^{7}\left (d x +c \right )\right )}{16 d a}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}\)	\(207\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^4*(1/16*I/(tan(d*x+c)-I)-5/12*I/(tan(d*x+c)-I)^3+1/8/(tan(d*x+c)-I)^4-7/16/(tan(d*x+c)-I)^2+1/32*ln(tan(
d*x+c)-I)-1/32*ln(tan(d*x+c)+I))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 54, normalized size = 0.43 \begin {gather*} \frac {{\left (24 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(24*I*d*x*e^(8*I*d*x + 8*I*c) + 24*e^(6*I*d*x + 6*I*c) - 8*e^(2*I*d*x + 2*I*c) + 3)*e^(-8*I*d*x - 8*I*c)
/(a^4*d)

________________________________________________________________________________________

Sympy [A]
time = 0.61, size = 156, normalized size = 1.24 \begin {gather*} \begin {cases} \frac {\left (6144 a^{8} d^{2} e^{14 i c} e^{- 2 i d x} - 2048 a^{8} d^{2} e^{10 i c} e^{- 6 i d x} + 768 a^{8} d^{2} e^{8 i c} e^{- 8 i d x}\right ) e^{- 16 i c}}{98304 a^{12} d^{3}} & \text {for}\: a^{12} d^{3} e^{16 i c} \neq 0 \\x \left (\frac {\left (i e^{8 i c} - 2 i e^{6 i c} + 2 i e^{2 i c} - i\right ) e^{- 8 i c}}{16 a^{4}} - \frac {i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {i x}{16 a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((6144*a**8*d**2*exp(14*I*c)*exp(-2*I*d*x) - 2048*a**8*d**2*exp(10*I*c)*exp(-6*I*d*x) + 768*a**8*d**
2*exp(8*I*c)*exp(-8*I*d*x))*exp(-16*I*c)/(98304*a**12*d**3), Ne(a**12*d**3*exp(16*I*c), 0)), (x*((I*exp(8*I*c)
- 2*I*exp(6*I*c) + 2*I*exp(2*I*c) - I)*exp(-8*I*c)/(16*a**4) - I/(16*a**4)), True)) + I*x/(16*a**4)

________________________________________________________________________________________

Giac [A]
time = 1.17, size = 88, normalized size = 0.70 \begin {gather*} -\frac {\frac {12 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {25 \, \tan \left (d x + c\right )^{4} - 124 i \, \tan \left (d x + c\right )^{3} - 54 \, \tan \left (d x + c\right )^{2} - 4 i \, \tan \left (d x + c\right ) - 7}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(12*log(tan(d*x + c) + I)/a^4 - 12*log(tan(d*x + c) - I)/a^4 + (25*tan(d*x + c)^4 - 124*I*tan(d*x + c)^
3 - 54*tan(d*x + c)^2 - 4*I*tan(d*x + c) - 7)/(a^4*(tan(d*x + c) - I)^4))/d

________________________________________________________________________________________

Mupad [B]
time = 3.90, size = 60, normalized size = 0.48 \begin {gather*} \frac {x\,1{}\mathrm {i}}{16\,a^4}+\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{16}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,13{}\mathrm {i}}{48}+\frac {1}{12}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*1i)/(16*a^4) + ((tan(c + d*x)*13i)/48 - tan(c + d*x)^2/4 + (tan(c + d*x)^3*1i)/16 + 1/12)/(a^4*d*(tan(c + d
*x)*1i + 1)^4)

________________________________________________________________________________________