3.1.23 \(\int \frac {1}{(c+d x)^3 (a+i a \tan (e+f x))} \, dx\) [23]

Optimal. Leaf size=227 \[ -\frac {i f}{2 a d^2 (c+d x)}-\frac {f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}+\frac {i f^2 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a d^3}+\frac {i f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}+\frac {f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))}+\frac {i f}{d^2 (c+d x) (a+i a \tan (e+f x))} \]

[Out]

-1/2*I*f/a/d^2/(d*x+c)-f^2*Ci(2*c*f/d+2*f*x)*cos(-2*e+2*c*f/d)/a/d^3+I*f^2*cos(-2*e+2*c*f/d)*Si(2*c*f/d+2*f*x)
/a/d^3-I*f^2*Ci(2*c*f/d+2*f*x)*sin(-2*e+2*c*f/d)/a/d^3-f^2*Si(2*c*f/d+2*f*x)*sin(-2*e+2*c*f/d)/a/d^3-1/2/d/(d*
x+c)^2/(a+I*a*tan(f*x+e))+I*f/d^2/(d*x+c)/(a+I*a*tan(f*x+e))

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Rubi [A]
time = 0.23, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3806, 3805, 3384, 3380, 3383} \begin {gather*} \frac {i f^2 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a d^3}-\frac {f^2 \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a d^3}+\frac {f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^3}+\frac {i f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^3}+\frac {i f}{d^2 (c+d x) (a+i a \tan (e+f x))}-\frac {i f}{2 a d^2 (c+d x)}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + d*x)^3*(a + I*a*Tan[e + f*x])),x]

[Out]

((-1/2*I)*f)/(a*d^2*(c + d*x)) - (f^2*Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/(a*d^3) + (I*f^2*Co
sIntegral[(2*c*f)/d + 2*f*x]*Sin[2*e - (2*c*f)/d])/(a*d^3) + (I*f^2*Cos[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d
+ 2*f*x])/(a*d^3) + (f^2*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(a*d^3) - 1/(2*d*(c + d*x)^2*(a
+ I*a*Tan[e + f*x])) + (I*f)/(d^2*(c + d*x)*(a + I*a*Tan[e + f*x]))

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3805

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

Rule 3806

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

Rubi steps

\begin {align*} \int \frac {1}{(c+d x)^3 (a+i a \tan (e+f x))} \, dx &=-\frac {i f}{2 a d^2 (c+d x)}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))}-\frac {(i f) \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))} \, dx}{d}\\ &=-\frac {i f}{2 a d^2 (c+d x)}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))}+\frac {i f}{d^2 (c+d x) (a+i a \tan (e+f x))}+\frac {\left (i f^2\right ) \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{a d^2}-\frac {f^2 \int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{a d^2}\\ &=-\frac {i f}{2 a d^2 (c+d x)}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))}+\frac {i f}{d^2 (c+d x) (a+i a \tan (e+f x))}+\frac {\left (i f^2 \cos \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{a d^2}-\frac {\left (f^2 \cos \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{a d^2}+\frac {\left (i f^2 \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{a d^2}+\frac {\left (f^2 \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{a d^2}\\ &=-\frac {i f}{2 a d^2 (c+d x)}-\frac {f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}+\frac {i f^2 \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a d^3}+\frac {i f^2 \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}+\frac {f^2 \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a d^3}-\frac {1}{2 d (c+d x)^2 (a+i a \tan (e+f x))}+\frac {i f}{d^2 (c+d x) (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.24, size = 285, normalized size = 1.26 \begin {gather*} \frac {\sec (e+f x) \left (\cos \left (\frac {c f}{d}\right )+i \sin \left (\frac {c f}{d}\right )\right ) \left (d \left (i d \cos \left (e+f \left (-\frac {c}{d}+x\right )\right )+(i d+2 c f+2 d f x) \cos \left (e+f \left (\frac {c}{d}+x\right )\right )-d \sin \left (e+f \left (-\frac {c}{d}+x\right )\right )+d \sin \left (e+f \left (\frac {c}{d}+x\right )\right )-2 i c f \sin \left (e+f \left (\frac {c}{d}+x\right )\right )-2 i d f x \sin \left (e+f \left (\frac {c}{d}+x\right )\right )\right )+4 f^2 (c+d x)^2 \text {CosIntegral}\left (\frac {2 f (c+d x)}{d}\right ) \left (i \cos \left (e-\frac {f (c+d x)}{d}\right )+\sin \left (e-\frac {f (c+d x)}{d}\right )\right )+4 f^2 (c+d x)^2 \left (\cos \left (e-\frac {f (c+d x)}{d}\right )-i \sin \left (e-\frac {f (c+d x)}{d}\right )\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )\right )}{4 a d^3 (c+d x)^2 (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c + d*x)^3*(a + I*a*Tan[e + f*x])),x]

[Out]

(Sec[e + f*x]*(Cos[(c*f)/d] + I*Sin[(c*f)/d])*(d*(I*d*Cos[e + f*(-(c/d) + x)] + (I*d + 2*c*f + 2*d*f*x)*Cos[e
+ f*(c/d + x)] - d*Sin[e + f*(-(c/d) + x)] + d*Sin[e + f*(c/d + x)] - (2*I)*c*f*Sin[e + f*(c/d + x)] - (2*I)*d
*f*x*Sin[e + f*(c/d + x)]) + 4*f^2*(c + d*x)^2*CosIntegral[(2*f*(c + d*x))/d]*(I*Cos[e - (f*(c + d*x))/d] + Si
n[e - (f*(c + d*x))/d]) + 4*f^2*(c + d*x)^2*(Cos[e - (f*(c + d*x))/d] - I*Sin[e - (f*(c + d*x))/d])*SinIntegra
l[(2*f*(c + d*x))/d]))/(4*a*d^3*(c + d*x)^2*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.45, size = 358, normalized size = 1.58

method	result	size
risch	\(-\frac {1}{4 d a \left (d x +c \right )^{2}}+\frac {i f^{3} {\mathrm e}^{-2 i \left (f x +e \right )} x}{2 a d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {f^{2} {\mathrm e}^{-2 i \left (f x +e \right )}}{4 a d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {i f^{3} {\mathrm e}^{-2 i \left (f x +e \right )} c}{2 a \,d^{2} \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {f^{2} {\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{a \,d^{3}}\)	\(216\)
default	\(\frac {-\frac {i f^{3} \left (-\frac {\sin \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right )^{2} d}+\frac {-\frac {2 \cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}-\frac {2 \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}}{d}\right )}{4}+\frac {f^{3} \left (-\frac {\cos \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 f x +2 e \right )}{\left (c f -d e +d \left (f x +e \right )\right ) d}+\frac {\frac {4 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}}{d}\right )}{4}-\frac {f^{3}}{4 \left (c f -d e +d \left (f x +e \right )\right )^{2} d}}{a f}\)	\(358\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x+c)^3/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/a/f*(-1/4*I*f^3*(-sin(2*f*x+2*e)/(c*f-d*e+d*(f*x+e))^2/d+(-2*cos(2*f*x+2*e)/(c*f-d*e+d*(f*x+e))/d-2*(2*Si(2*
f*x+2*e+2*(c*f-d*e)/d)*cos(2*(c*f-d*e)/d)/d-2*Ci(2*f*x+2*e+2*(c*f-d*e)/d)*sin(2*(c*f-d*e)/d)/d)/d)/d)+1/4*f^3*
(-cos(2*f*x+2*e)/(c*f-d*e+d*(f*x+e))^2/d-(-2*sin(2*f*x+2*e)/(c*f-d*e+d*(f*x+e))/d+2*(2*Si(2*f*x+2*e+2*(c*f-d*e
)/d)*sin(2*(c*f-d*e)/d)/d+2*Ci(2*f*x+2*e+2*(c*f-d*e)/d)*cos(2*(c*f-d*e)/d)/d)/d)/d)-1/4*f^3/(c*f-d*e+d*(f*x+e)
)^2/d)

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Maxima [A]
time = 0.42, size = 167, normalized size = 0.74 \begin {gather*} -\frac {2 \, f^{3} \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) E_{3}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + 2 i \, f^{3} E_{3}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) \sin \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) + f^{3}}{4 \, {\left ({\left (f x + e\right )}^{2} a d^{3} + a c^{2} d f^{2} - 2 \, a c d^{2} f e + a d^{3} e^{2} + 2 \, {\left (a c d^{2} f - a d^{3} e\right )} {\left (f x + e\right )}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)^3/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

-1/4*(2*f^3*cos(2*(c*f - d*e)/d)*exp_integral_e(3, -2*(-I*(f*x + e)*d - I*c*f + I*d*e)/d) + 2*I*f^3*exp_integr
al_e(3, -2*(-I*(f*x + e)*d - I*c*f + I*d*e)/d)*sin(2*(c*f - d*e)/d) + f^3)/(((f*x + e)^2*a*d^3 + a*c^2*d*f^2 -
2*a*c*d^2*f*e + a*d^3*e^2 + 2*(a*c*d^2*f - a*d^3*e)*(f*x + e))*f)

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Fricas [A]
time = 0.37, size = 133, normalized size = 0.59 \begin {gather*} \frac {{\left (2 i \, d^{2} f x + 2 i \, c d f - d^{2} - {\left (4 \, {\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x + c^{2} f^{2}\right )} {\rm Ei}\left (-\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (-i \, c f + i \, d e\right )}}{d}\right )} + d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, {\left (a d^{5} x^{2} + 2 \, a c d^{4} x + a c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)^3/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/4*(2*I*d^2*f*x + 2*I*c*d*f - d^2 - (4*(d^2*f^2*x^2 + 2*c*d*f^2*x + c^2*f^2)*Ei(-2*(I*d*f*x + I*c*f)/d)*e^(-2
*(-I*c*f + I*d*e)/d) + d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*d^5*x^2 + 2*a*c*d^4*x + a*c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{c^{3} \tan {\left (e + f x \right )} - i c^{3} + 3 c^{2} d x \tan {\left (e + f x \right )} - 3 i c^{2} d x + 3 c d^{2} x^{2} \tan {\left (e + f x \right )} - 3 i c d^{2} x^{2} + d^{3} x^{3} \tan {\left (e + f x \right )} - i d^{3} x^{3}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)**3/(a+I*a*tan(f*x+e)),x)

[Out]

-I*Integral(1/(c**3*tan(e + f*x) - I*c**3 + 3*c**2*d*x*tan(e + f*x) - 3*I*c**2*d*x + 3*c*d**2*x**2*tan(e + f*x
) - 3*I*c*d**2*x**2 + d**3*x**3*tan(e + f*x) - I*d**3*x**3), x)/a

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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. 540 vs. \(2 (218) = 436\).
time = 0.52, size = 540, normalized size = 2.38 \begin {gather*} -\frac {4 \, d^{2} f^{2} x^{2} \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 i \, d^{2} f^{2} x^{2} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) - 4 i \, d^{2} f^{2} x^{2} \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, d^{2} f^{2} x^{2} \sin \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 \, c d f^{2} x \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 i \, c d f^{2} x \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) - 8 i \, c d f^{2} x \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 8 \, c d f^{2} x \sin \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, c^{2} f^{2} \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 i \, c^{2} f^{2} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) - 4 i \, c^{2} f^{2} \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 4 \, c^{2} f^{2} \sin \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 2 i \, d^{2} f x \cos \left (2 \, f x\right ) - 2 \, d^{2} f x \sin \left (2 \, f x\right ) - 2 i \, c d f \cos \left (2 \, f x\right ) - 2 \, c d f \sin \left (2 \, f x\right ) + d^{2} \cos \left (2 \, f x\right ) + d^{2} \cos \left (2 \, e\right ) - i \, d^{2} \sin \left (2 \, f x\right ) + i \, d^{2} \sin \left (2 \, e\right )}{4 \, {\left (a d^{5} x^{2} \cos \left (2 \, e\right ) + i \, a d^{5} x^{2} \sin \left (2 \, e\right ) + 2 \, a c d^{4} x \cos \left (2 \, e\right ) + 2 i \, a c d^{4} x \sin \left (2 \, e\right ) + a c^{2} d^{3} \cos \left (2 \, e\right ) + i \, a c^{2} d^{3} \sin \left (2 \, e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)^3/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/4*(4*d^2*f^2*x^2*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/d) + 4*I*d^2*f^2*x^2*cos_integral(-2*(d*f*x + c
*f)/d)*sin(2*c*f/d) - 4*I*d^2*f^2*x^2*cos(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + 4*d^2*f^2*x^2*sin(2*c*f/d
)*sin_integral(2*(d*f*x + c*f)/d) + 8*c*d*f^2*x*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/d) + 8*I*c*d*f^2*x*
cos_integral(-2*(d*f*x + c*f)/d)*sin(2*c*f/d) - 8*I*c*d*f^2*x*cos(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + 8
*c*d*f^2*x*sin(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + 4*c^2*f^2*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)
/d) + 4*I*c^2*f^2*cos_integral(-2*(d*f*x + c*f)/d)*sin(2*c*f/d) - 4*I*c^2*f^2*cos(2*c*f/d)*sin_integral(2*(d*f
*x + c*f)/d) + 4*c^2*f^2*sin(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) - 2*I*d^2*f*x*cos(2*f*x) - 2*d^2*f*x*sin
(2*f*x) - 2*I*c*d*f*cos(2*f*x) - 2*c*d*f*sin(2*f*x) + d^2*cos(2*f*x) + d^2*cos(2*e) - I*d^2*sin(2*f*x) + I*d^2
*sin(2*e))/(a*d^5*x^2*cos(2*e) + I*a*d^5*x^2*sin(2*e) + 2*a*c*d^4*x*cos(2*e) + 2*I*a*c*d^4*x*sin(2*e) + a*c^2*
d^3*cos(2*e) + I*a*c^2*d^3*sin(2*e))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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