3.1.27 \(\int \frac {1}{(c+d x) (a+i a \tan (e+f x))^2} \, dx\) [27]

Optimal. Leaf size=305 \[ \frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\cos \left (4 e-\frac {4 c f}{d}\right ) \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}-\frac {i \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.55, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3809, 3384, 3380, 3383, 3393} \begin {gather*} -\frac {i \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}+\frac {\text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{4 a^2 d}+\frac {\log (c+d x)}{4 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + d*x)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

(Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) + (Cos[4*e - (4*c*f)/d]*CosIntegral[(4*c*f)/d
+ 4*f*x])/(4*a^2*d) + Log[c + d*x]/(4*a^2*d) - ((I/4)*CosIntegral[(4*c*f)/d + 4*f*x]*Sin[4*e - (4*c*f)/d])/(a^
2*d) - ((I/2)*CosIntegral[(2*c*f)/d + 2*f*x]*Sin[2*e - (2*c*f)/d])/(a^2*d) - ((I/2)*Cos[2*e - (2*c*f)/d]*SinIn
tegral[(2*c*f)/d + 2*f*x])/(a^2*d) - (Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(2*a^2*d) - ((I/4)*
Cos[4*e - (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(a^2*d) - (Sin[4*e - (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4
*f*x])/(4*a^2*d)

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3809

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

Rubi steps

\begin {align*} \int \frac {1}{(c+d x) (a+i a \tan (e+f x))^2} \, dx &=\int \left (\frac {1}{4 a^2 (c+d x)}+\frac {\cos (2 e+2 f x)}{2 a^2 (c+d x)}+\frac {\cos ^2(2 e+2 f x)}{4 a^2 (c+d x)}-\frac {i \sin (2 e+2 f x)}{2 a^2 (c+d x)}-\frac {\sin ^2(2 e+2 f x)}{4 a^2 (c+d x)}-\frac {i \sin (4 e+4 f x)}{4 a^2 (c+d x)}\right ) \, dx\\ &=\frac {\log (c+d x)}{4 a^2 d}-\frac {i \int \frac {\sin (4 e+4 f x)}{c+d x} \, dx}{4 a^2}-\frac {i \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{2 a^2}+\frac {\int \frac {\cos ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}-\frac {\int \frac {\sin ^2(2 e+2 f x)}{c+d x} \, dx}{4 a^2}+\frac {\int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{2 a^2}\\ &=\frac {\log (c+d x)}{4 a^2 d}-\frac {\int \left (\frac {1}{2 (c+d x)}-\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}+\frac {\int \left (\frac {1}{2 (c+d x)}+\frac {\cos (4 e+4 f x)}{2 (c+d x)}\right ) \, dx}{4 a^2}-\frac {\left (i \cos \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\sin \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{4 a^2}-\frac {\left (i \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}{2 a^2}+\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\cos \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}-\frac {\left (i \sin \left (4 e-\frac {4 c f}{d}\right )\right ) \int \frac {\cos \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{4 a^2}-\frac {\left (i \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}{2 a^2}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \int \frac {\sin \left (\frac {2 c f}{d}+2 f x\right )}{c+d x} \, dx}{2 a^2}\\ &=\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \frac {\int \frac {\cos (4 e+4 f x)}{c+d x} \, dx}{8 a^2}\\ &=\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac {\cos \left (4 e-\frac {4 c f}{d}\right ) \int \frac {\cos \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \int \frac {\sin \left (\frac {4 c f}{d}+4 f x\right )}{c+d x} \, dx}{8 a^2}\right )\\ &=\frac {\cos \left (2 e-\frac {2 c f}{d}\right ) \text {Ci}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}+\frac {\log (c+d x)}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{4 a^2 d}-\frac {i \text {Ci}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a^2 d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{2 a^2 d}-\frac {i \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{4 a^2 d}+2 \left (\frac {\cos \left (4 e-\frac {4 c f}{d}\right ) \text {Ci}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^2 d}-\frac {\sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{8 a^2 d}\right )\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 211, normalized size = 0.69 \begin {gather*} \frac {\left (\cos \left (2 e-\frac {2 c f}{d}\right )-i \sin \left (2 e-\frac {2 c f}{d}\right )\right ) \left (2 \text {CosIntegral}\left (\frac {2 f (c+d x)}{d}\right )+\cos \left (2 e-\frac {2 c f}{d}\right ) \log (f (c+d x))+\text {CosIntegral}\left (\frac {4 f (c+d x)}{d}\right ) \left (\cos \left (2 e-\frac {2 c f}{d}\right )-i \sin \left (2 e-\frac {2 c f}{d}\right )\right )+i \log (f (c+d x)) \sin \left (2 e-\frac {2 c f}{d}\right )-2 i \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right )}{4 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c + d*x)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

((Cos[2*e - (2*c*f)/d] - I*Sin[2*e - (2*c*f)/d])*(2*CosIntegral[(2*f*(c + d*x))/d] + Cos[2*e - (2*c*f)/d]*Log[
f*(c + d*x)] + CosIntegral[(4*f*(c + d*x))/d]*(Cos[2*e - (2*c*f)/d] - I*Sin[2*e - (2*c*f)/d]) + I*Log[f*(c + d
*x)]*Sin[2*e - (2*c*f)/d] - (2*I)*SinIntegral[(2*f*(c + d*x))/d] - I*Cos[2*e - (2*c*f)/d]*SinIntegral[(4*f*(c
+ d*x))/d] - Sin[2*e - (2*c*f)/d]*SinIntegral[(4*f*(c + d*x))/d]))/(4*a^2*d)

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Maple [A]
time = 0.44, size = 377, normalized size = 1.24

method	result	size
risch	\(\frac {\ln \left (d x +c \right )}{4 a^{2} d}-\frac {{\mathrm e}^{\frac {4 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 4 i f x +4 i e +\frac {4 i \left (c f -d e \right )}{d}\right )}{4 a^{2} d}-\frac {{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{2 a^{2} d}\)	\(114\)
default	\(\frac {-\frac {i f \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{4}-\frac {i f \left (\frac {4 \sinIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}-\frac {4 \cosineIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}\right )}{16}+\frac {f \left (\frac {4 \sinIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}+\frac {4 \cosineIntegral \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}\right )}{16}+\frac {f \left (\frac {2 \sinIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {2 \cosineIntegral \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{4}+\frac {f \ln \left (c f -d e +d \left (f x +e \right )\right )}{4 d}}{a^{2} f}\)	\(377\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x+c)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/a^2/f*(-1/4*I*f*(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)-1/16*I*f*(4*Si(4*f*x+4*e+4*(c*f-d*e)/d)*cos(4*(c*f-d*e)/d)/d-4*Ci(4*f*x+4*e+4*(c*f-d*e)/d)*sin(4*(
c*f-d*e)/d)/d)+1/16*f*(4*Si(4*f*x+4*e+4*(c*f-d*e)/d)*sin(4*(c*f-d*e)/d)/d+4*Ci(4*f*x+4*e+4*(c*f-d*e)/d)*cos(4*
(c*f-d*e)/d)/d)+1/4*f*(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)+1/4*f*ln(c*f-d*e+d*(f*x+e))/d)

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Maxima [A]
time = 0.36, size = 208, normalized size = 0.68 \begin {gather*} -\frac {2 \, f \cos \left (\frac {2 \, {\left (c f - d e\right )}}{d}\right ) E_{1}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + f \cos \left (\frac {4 \, {\left (c f - d e\right )}}{d}\right ) E_{1}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) + i \, f E_{1}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d - i \, c f + i \, d e\right )}}{d}\right ) \sin \left (\frac {4 \, {\left (c f - d e\right )}}{d}\right ) + 2 i \, f E_{1}\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 \log \left ({\left (f x + e\right )} d + c f - d e\right )}{4 \, a^{2} d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 86, normalized size = 0.28 \begin {gather*} \frac {2 \, {\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 )} + {\rm Ei}\left (-\frac {4 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {4 \, {\left (-i \, c f + i \, d e\right )}}{d}\right )} + \log \left (\frac {d x + c}{d}\right )}{4 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.55, size = 420, normalized size = 1.38 \begin {gather*} \frac {2 \, \cos \left (\frac {2 \, c f}{d}\right ) \cos \left (2 \, e\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + \cos \left (2 \, e\right )^{2} \log \left (d x + c\right ) + 2 i \, \cos \left (2 \, e\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) + 2 i \, \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (2 \, e\right ) + 2 i \, \cos \left (2 \, e\right ) \log \left (d x + c\right ) \sin \left (2 \, e\right ) - 2 \, \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) \sin \left (2 \, e\right ) - \log \left (d x + c\right ) \sin \left (2 \, e\right )^{2} - 2 i \, \cos \left (\frac {2 \, c f}{d}\right ) \cos \left (2 \, e\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 2 \, \cos \left (2 \, e\right ) \sin \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 2 \, \cos \left (\frac {2 \, c f}{d}\right ) \sin \left (2 \, e\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + 2 i \, \sin \left (\frac {2 \, c f}{d}\right ) \sin \left (2 \, e\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + \cos \left (\frac {4 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) + i \, \operatorname {Ci}\left (-\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {4 \, c f}{d}\right ) - i \, \cos \left (\frac {4 \, c f}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (d f x + c f\right )}}{d}\right ) + \sin \left (\frac {4 \, c f}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (d f x + c f\right )}}{d}\right )}{4 \, {\left (a^{2} d \cos \left (2 \, e\right )^{2} + 2 i \, a^{2} d \cos \left (2 \, e\right ) \sin \left (2 \, e\right ) - a^{2} d \sin \left (2 \, e\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x+c)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

1/4*(2*cos(2*c*f/d)*cos(2*e)*cos_integral(-2*(d*f*x + c*f)/d) + cos(2*e)^2*log(d*x + c) + 2*I*cos(2*e)*cos_int
egral(-2*(d*f*x + c*f)/d)*sin(2*c*f/d) + 2*I*cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/d)*sin(2*e) + 2*I*cos(
2*e)*log(d*x + c)*sin(2*e) - 2*cos_integral(-2*(d*f*x + c*f)/d)*sin(2*c*f/d)*sin(2*e) - log(d*x + c)*sin(2*e)^
2 - 2*I*cos(2*c*f/d)*cos(2*e)*sin_integral(2*(d*f*x + c*f)/d) + 2*cos(2*e)*sin(2*c*f/d)*sin_integral(2*(d*f*x
+ c*f)/d) + 2*cos(2*c*f/d)*sin(2*e)*sin_integral(2*(d*f*x + c*f)/d) + 2*I*sin(2*c*f/d)*sin(2*e)*sin_integral(2
*(d*f*x + c*f)/d) + cos(4*c*f/d)*cos_integral(-4*(d*f*x + c*f)/d) + I*cos_integral(-4*(d*f*x + c*f)/d)*sin(4*c
*f/d) - I*cos(4*c*f/d)*sin_integral(4*(d*f*x + c*f)/d) + sin(4*c*f/d)*sin_integral(4*(d*f*x + c*f)/d))/(a^2*d*
cos(2*e)^2 + 2*I*a^2*d*cos(2*e)*sin(2*e) - a^2*d*sin(2*e)^2)

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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 )}^2\,\left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^2*(c + d*x)),x)

[Out]

int(1/((a + a*tan(e + f*x)*1i)^2*(c + d*x)), x)

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