3.1.21 \(\int \frac {1}{(c+d x) (a+i a \tan (e+f x))} \, dx\) [21]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.21, 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 = {3807, 3384, 3380, 3383} \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 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} \end {gather*}

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 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 3807

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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.38, size = 166, normalized size = 1.03 \begin {gather*} \frac {\sec (e+f x) \left (-i \cos \left (f \left (\frac {c}{d}+x\right )\right )+\sin \left (f \left (\frac {c}{d}+x\right )\right )\right ) \left (\text {CosIntegral}\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 )+\log (f (c+d x)) \left (\cos \left (e-\frac {c f}{d}\right )+i \sin \left (e-\frac {c f}{d}\right )\right )+\left (-i \cos \left (e-\frac {c f}{d}\right )-\sin \left (e-\frac {c f}{d}\right )\right ) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )\right )}{2 a d (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[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]))

________________________________________________________________________________________

Maple [A]
time = 0.46, size = 65, normalized size = 0.40

method	result	size
risch	\(\frac {\ln \left (d x +c \right )}{2 d a}-\frac {{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \expIntegral \left (1, 2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{2 a d}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(d*x+c)/d/a-1/2/a/d*exp(2*I*(c*f-d*e)/d)*Ei(1,2*I*f*x+2*I*e+2*I*(c*f-d*e)/d)

________________________________________________________________________________________

Maxima [A]
time = 0.36, size = 120, normalized size = 0.75 \begin {gather*} -\frac {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 ) + 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 )}{2 \, a 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)),x, algorithm="maxima")

[Out]

-1/2*(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) + 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*d*f)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 52, normalized size = 0.32 \begin {gather*} \frac {{\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 )} + \log \left (\frac {d x + c}{d}\right )}{2 \, a 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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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} \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)),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.53, size = 142, normalized size = 0.88 \begin {gather*} \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 )}} \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)),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))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________