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

Optimal. Leaf size=168 \[ -\frac {f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a d^2}-\frac {i f \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^2}-\frac {1}{d (c+d x) (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3724, 3303, 3299, 3302} \[ -\frac {f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a d^2}-\frac {i f \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a d^2}-\frac {1}{d (c+d x) (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3724

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]

Rubi steps

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

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Mathematica [A]  time = 0.85, size = 224, normalized size = 1.33 \[ \frac {\sec (e+f x) \left (\cos \left (\frac {c f}{d}\right )+i \sin \left (\frac {c f}{d}\right )\right ) \left (-2 f (c+d x) \text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cos \left (e-\frac {f (c+d x)}{d}\right )-i \sin \left (e-\frac {f (c+d x)}{d}\right )\right )+2 f (c+d x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right ) \left (\sin \left (e-\frac {f (c+d x)}{d}\right )+i \cos \left (e-\frac {f (c+d x)}{d}\right )\right )+d \left (-\sin \left (f \left (x-\frac {c}{d}\right )+e\right )+\sin \left (f \left (\frac {c}{d}+x\right )+e\right )+i \cos \left (f \left (x-\frac {c}{d}\right )+e\right )+i \cos \left (f \left (\frac {c}{d}+x\right )+e\right )\right )\right )}{2 a d^2 (c+d x) (\tan (e+f x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[(c*f)/d] + I*Sin[(c*f)/d])*(d*(I*Cos[e + f*(-(c/d) + x)] + I*Cos[e + f*(c/d + x)] - Sin[e +
f*(-(c/d) + x)] + Sin[e + f*(c/d + x)]) - 2*f*(c + d*x)*CosIntegral[(2*f*(c + d*x))/d]*(Cos[e - (f*(c + d*x))
/d] - I*Sin[e - (f*(c + d*x))/d]) + 2*f*(c + d*x)*(I*Cos[e - (f*(c + d*x))/d] + Sin[e - (f*(c + d*x))/d])*SinI
ntegral[(2*f*(c + d*x))/d]))/(2*a*d^2*(c + d*x)*(-I + Tan[e + f*x]))

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fricas [A]  time = 0.58, size = 84, normalized size = 0.50 \[ \frac {{\left ({\left ({\left (-2 i \, d f x - 2 i \, c f\right )} {\rm Ei}\left (\frac {-2 i \, d f x - 2 i \, c f}{d}\right ) e^{\left (\frac {-2 i \, d e + 2 i \, c f}{d}\right )} - d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (a d^{3} x + a c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 74.99, size = 1099, normalized size = 6.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.47, size = 285, normalized size = 1.70 \[ \frac {-\frac {i f^{2} \left (-\frac {2 \sin \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {4 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}+\frac {4 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}}{d}\right )}{4}+\frac {f^{2} \left (-\frac {2 \cos \left (2 f x +2 e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{d}-\frac {2 \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{d}\right )}{d}\right )}{4}-\frac {f^{2}}{2 \left (\left (f x +e \right ) d +c f -d e \right ) d}}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 117, normalized size = 0.70 \[ -\frac {f^{2} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + i \, f^{2} E_{2}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + f^{2}}{2 \, {\left ({\left (f x + e\right )} a d^{2} - a d^{2} e + a c d f\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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) + I*f^2*exp_integra
l_e(2, (2*I*(f*x + e)*d - 2*I*d*e + 2*I*c*f)/d)*sin(-2*(d*e - c*f)/d) + f^2)/(((f*x + e)*a*d^2 - a*d^2*e + a*c
*d*f)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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