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

Optimal. Leaf size=161 \[ -\frac {i \text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a d}+\frac {\text {Ci}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 a d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}+\frac {\log (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 161, 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 = {3726, 3303, 3299, 3302} \[ -\frac {i \text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{2 a d}+\frac {\text {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{2 a d}-\frac {\sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}-\frac {i \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{2 a d}+\frac {\log (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3726

Int[1/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Log[c + d*x]/(2*a*d), x
] + (Dist[1/(2*a), Int[Cos[2*e + 2*f*x]/(c + d*x), x], x] + Dist[1/(2*b), Int[Sin[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) (a+i a \tan (e+f x))} \, dx &=\frac {\log (c+d x)}{2 a d}-\frac {i \int \frac {\sin (2 e+2 f x)}{c+d x} \, dx}{2 a}+\frac {\int \frac {\cos (2 e+2 f x)}{c+d x} \, dx}{2 a}\\ &=\frac {\log (c+d x)}{2 a d}-\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}+\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}-\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}-\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}\\ &=\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 d}+\frac {\log (c+d x)}{2 a 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 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 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 d}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 166, normalized size = 1.03 \[ \frac {\sec (e+f x) \left (\sin \left (f \left (\frac {c}{d}+x\right )\right )-i \cos \left (f \left (\frac {c}{d}+x\right )\right )\right ) \left (\text {Ci}\left (\frac {2 f (c+d x)}{d}\right ) \left (\cos \left (e-\frac {c f}{d}\right )-i \sin \left (e-\frac {c f}{d}\right )\right )+\text {Si}\left (\frac {2 f (c+d x)}{d}\right ) \left (-\sin \left (e-\frac {c f}{d}\right )-i \cos \left (e-\frac {c f}{d}\right )\right )+\log (f (c+d x)) \left (\cos \left (e-\frac {c f}{d}\right )+i \sin \left (e-\frac {c f}{d}\right )\right )\right )}{2 a d (\tan (e+f x)-i)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 49, normalized size = 0.30 \[ \frac {{\rm Ei}\left (\frac {-2 i \, d f x - 2 i \, c f}{d}\right ) e^{\left (\frac {-2 i \, d e + 2 i \, c f}{d}\right )} + \log \left (\frac {d x + c}{d}\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.78, size = 142, normalized size = 0.88 \[ \frac {\cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + \cos \left (2 \, e\right ) \log \left (d x + c\right ) + i \, \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) \sin \left (\frac {2 \, c f}{d}\right ) + i \, \log \left (d x + c\right ) \sin \left (2 \, e\right ) - i \, \cos \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + \sin \left (\frac {2 \, c f}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + c f\right )}}{d}\right )}{2 \, {\left (a d \cos \left (2 \, e\right ) + i \, a d \sin \left (2 \, e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(cos(2*c*f/d)*cos_integral(-2*(d*f*x + c*f)/d) + cos(2*e)*log(d*x + c) + I*cos_integral(-2*(d*f*x + c*f)/d
)*sin(2*c*f/d) + I*log(d*x + c)*sin(2*e) - I*cos(2*c*f/d)*sin_integral(2*(d*f*x + c*f)/d) + sin(2*c*f/d)*sin_i
ntegral(2*(d*f*x + c*f)/d))/(a*d*cos(2*e) + I*a*d*sin(2*e))

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maple [A]  time = 0.51, size = 204, normalized size = 1.27 \[ -\frac {i \Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{2 a d}+\frac {i \Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{2 a d}+\frac {\Si \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \sin \left (\frac {2 c f -2 d e}{d}\right )}{2 a d}+\frac {\Ci \left (2 f x +2 e +\frac {2 c f -2 d e}{d}\right ) \cos \left (\frac {2 c f -2 d e}{d}\right )}{2 a d}+\frac {\ln \left (\left (f x +e \right ) d +c f -d e \right )}{2 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.79, size = 110, normalized size = 0.68 \[ -\frac {f \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{1}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) + i \, f E_{1}\left (\frac {2 i \, {\left (f x + e\right )} d - 2 i \, d e + 2 i \, c f}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) - f \log \left ({\left (f x + e\right )} d - d e + c f\right )}{2 \, a d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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