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

Optimal. Leaf size=436 \[ -\frac {f \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {Ci}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.74, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3728, 3297, 3303, 3299, 3302, 3313, 12} \[ -\frac {f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \text {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*a^2*d*(c + d*x)) - Cos[2*e + 2*f*x]/(2*a^2*d*(c + d*x)) - Cos[2*e + 2*f*x]^2/(4*a^2*d*(c + d*x)) - (I*f*
Cos[2*e - (2*c*f)/d]*CosIntegral[(2*c*f)/d + 2*f*x])/(a^2*d^2) - (I*f*Cos[4*e - (4*c*f)/d]*CosIntegral[(4*c*f)
/d + 4*f*x])/(a^2*d^2) - (f*CosIntegral[(4*c*f)/d + 4*f*x]*Sin[4*e - (4*c*f)/d])/(a^2*d^2) - (f*CosIntegral[(2
*c*f)/d + 2*f*x]*Sin[2*e - (2*c*f)/d])/(a^2*d^2) + ((I/2)*Sin[2*e + 2*f*x])/(a^2*d*(c + d*x)) + Sin[2*e + 2*f*
x]^2/(4*a^2*d*(c + d*x)) + ((I/4)*Sin[4*e + 4*f*x])/(a^2*d*(c + d*x)) - (f*Cos[2*e - (2*c*f)/d]*SinIntegral[(2
*c*f)/d + 2*f*x])/(a^2*d^2) + (I*f*Sin[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(a^2*d^2) - (f*Cos[4*e
- (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(a^2*d^2) + (I*f*Sin[4*e - (4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*
f*x])/(a^2*d^2)

Rule 12

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

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3299

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 3302

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 3303

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 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 3728

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

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Mathematica [A]  time = 1.61, size = 467, normalized size = 1.07 \[ -\frac {\left (\cos \left (2 \left (f \left (x-\frac {c}{d}\right )+e\right )\right )-i \sin \left (2 \left (f \left (x-\frac {c}{d}\right )+e\right )\right )\right ) \left (4 f (c+d x) \text {Ci}\left (\frac {4 f (c+d x)}{d}\right ) \left (\sin \left (2 e-\frac {2 f (c+d x)}{d}\right )+i \cos \left (2 e-\frac {2 f (c+d x)}{d}\right )\right )+4 i f (c+d x) (\cos (2 f x)+i \sin (2 f x)) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right )-4 i c f \text {Si}\left (\frac {4 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 f (c+d x)}{d}\right )-4 i d f x \text {Si}\left (\frac {4 f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 f (c+d x)}{d}\right )+4 c f \text {Si}\left (\frac {4 f (c+d x)}{d}\right ) \cos \left (2 e-\frac {2 f (c+d x)}{d}\right )+4 d f x \text {Si}\left (\frac {4 f (c+d x)}{d}\right ) \cos \left (2 e-\frac {2 f (c+d x)}{d}\right )+i d \sin \left (2 \left (f \left (x-\frac {c}{d}\right )+e\right )\right )-i d \sin \left (2 \left (f \left (\frac {c}{d}+x\right )+e\right )\right )+d \cos \left (2 \left (f \left (x-\frac {c}{d}\right )+e\right )\right )+d \cos \left (2 \left (f \left (\frac {c}{d}+x\right )+e\right )\right )+4 i c f \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 i d f x \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 c f \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 d f x \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )-2 i d \sin \left (\frac {2 c f}{d}\right )+2 d \cos \left (\frac {2 c f}{d}\right )\right )}{4 a^2 d^2 (c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*((Cos[2*(e + f*(-(c/d) + x))] - I*Sin[2*(e + f*(-(c/d) + x))])*(2*d*Cos[(2*c*f)/d] + d*Cos[2*(e + f*(-(c/
d) + x))] + d*Cos[2*(e + f*(c/d + x))] - (2*I)*d*Sin[(2*c*f)/d] + (4*I)*f*(c + d*x)*CosIntegral[(2*f*(c + d*x)
)/d]*(Cos[2*f*x] + I*Sin[2*f*x]) + I*d*Sin[2*(e + f*(-(c/d) + x))] - I*d*Sin[2*(e + f*(c/d + x))] + 4*f*(c + d
*x)*CosIntegral[(4*f*(c + d*x))/d]*(I*Cos[2*e - (2*f*(c + d*x))/d] + Sin[2*e - (2*f*(c + d*x))/d]) + 4*c*f*Cos
[2*f*x]*SinIntegral[(2*f*(c + d*x))/d] + 4*d*f*x*Cos[2*f*x]*SinIntegral[(2*f*(c + d*x))/d] + (4*I)*c*f*Sin[2*f
*x]*SinIntegral[(2*f*(c + d*x))/d] + (4*I)*d*f*x*Sin[2*f*x]*SinIntegral[(2*f*(c + d*x))/d] + 4*c*f*Cos[2*e - (
2*f*(c + d*x))/d]*SinIntegral[(4*f*(c + d*x))/d] + 4*d*f*x*Cos[2*e - (2*f*(c + d*x))/d]*SinIntegral[(4*f*(c +
d*x))/d] - (4*I)*c*f*Sin[2*e - (2*f*(c + d*x))/d]*SinIntegral[(4*f*(c + d*x))/d] - (4*I)*d*f*x*Sin[2*e - (2*f*
(c + d*x))/d]*SinIntegral[(4*f*(c + d*x))/d]))/(a^2*d^2*(c + d*x))

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fricas [A]  time = 0.44, size = 140, normalized size = 0.32 \[ \frac {{\left ({\left ({\left (-4 i \, d f x - 4 i \, c f\right )} {\rm Ei}\left (\frac {-2 i \, d f x - 2 i \, c f}{d}\right ) e^{\left (\frac {-2 i \, d e + 2 i \, c f}{d}\right )} + {\left (-4 i \, d f x - 4 i \, c f\right )} {\rm Ei}\left (\frac {-4 i \, d f x - 4 i \, c f}{d}\right ) e^{\left (\frac {-4 i \, d e + 4 i \, c f}{d}\right )} - d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d e^{\left (2 i \, f x + 2 i \, e\right )} - d\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.57, size = 536, normalized size = 1.23 \[ \frac {-\frac {i f^{2} \left (-\frac {2 \sin \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {4 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}\right )}{4}-\frac {i f^{2} \left (-\frac {4 \sin \left (4 f x +4 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {16 \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}+\frac {16 \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}}{d}\right )}{16}+\frac {f^{2} \left (-\frac {4 \cos \left (4 f x +4 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {4 \left (\frac {4 \Si \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \cos \left (\frac {4 c f -4 d e}{d}\right )}{d}-\frac {4 \Ci \left (4 f x +4 e +\frac {4 c f -4 d e}{d}\right ) \sin \left (\frac {4 c f -4 d e}{d}\right )}{d}\right )}{d}\right )}{16}+\frac {f^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}-\frac {f^{2}}{4 \left (\left (f x +e \right ) d +c f -d e \right ) d}}{a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 210, normalized size = 0.48 \[ -\frac {64 \, f^{2} \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) + 128 \, f^{2} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + 128 i \, f^{2} E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + 64 i \, f^{2} E_{2}\left (\frac {4 i \, {\left (f x + e\right )} d - 4 i \, d e + 4 i \, c f}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + 64 \, f^{2}}{256 \, {\left ({\left (f x + e\right )} a^{2} d^{2} - a^{2} d^{2} e + a^{2} c d f\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(64*f^2*cos(-4*(d*e - c*f)/d)*exp_integral_e(2, (4*I*(f*x + e)*d - 4*I*d*e + 4*I*c*f)/d) + 128*f^2*cos(
-2*(d*e - c*f)/d)*exp_integral_e(2, (2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d) + 128*I*f^2*exp_integral_e(2, (2*
I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d)*sin(-2*(d*e - c*f)/d) + 64*I*f^2*exp_integral_e(2, (4*I*(f*x + e)*d - 4*
I*d*e + 4*I*c*f)/d)*sin(-4*(d*e - c*f)/d) + 64*f^2)/(((f*x + e)*a^2*d^2 - a^2*d^2*e + a^2*c*d*f)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{c^{2} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \tan {\left (e + f x \right )} - c^{2} + 2 c d x \tan ^{2}{\left (e + f x \right )} - 4 i c d x \tan {\left (e + f x \right )} - 2 c d x + d^{2} x^{2} \tan ^{2}{\left (e + f x \right )} - 2 i d^{2} x^{2} \tan {\left (e + f x \right )} - d^{2} x^{2}}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(c**2*tan(e + f*x)**2 - 2*I*c**2*tan(e + f*x) - c**2 + 2*c*d*x*tan(e + f*x)**2 - 4*I*c*d*x*tan(e +
f*x) - 2*c*d*x + d**2*x**2*tan(e + f*x)**2 - 2*I*d**2*x**2*tan(e + f*x) - d**2*x**2), x)/a**2

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