3.1.20 \(\int \frac {c+d x}{a+i a \tan (e+f x)} \, dx\) [20]

Optimal. Leaf size=84 \[ -\frac {i d x}{4 a f}+\frac {(c+d x)^2}{4 a d}+\frac {d}{4 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3804, 3560, 8} \begin {gather*} \frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}+\frac {d}{4 f^2 (a+i a \tan (e+f x))}-\frac {i d x}{4 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3804

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

Rubi steps

\begin {align*} \int \frac {c+d x}{a+i a \tan (e+f x)} \, dx &=\frac {(c+d x)^2}{4 a d}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}-\frac {(i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{2 f}\\ &=\frac {(c+d x)^2}{4 a d}+\frac {d}{4 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}-\frac {(i d) \int 1 \, dx}{4 a f}\\ &=-\frac {i d x}{4 a f}+\frac {(c+d x)^2}{4 a d}+\frac {d}{4 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 96, normalized size = 1.14 \begin {gather*} \frac {-i \left (2 c f (i+2 f x)+d \left (1+2 i f x+2 f^2 x^2\right )\right )+\left (2 c f (-i+2 f x)+d \left (-1-2 i f x+2 f^2 x^2\right )\right ) \tan (e+f x)}{8 a f^2 (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(2*c*f*(I + 2*f*x) + d*(1 + (2*I)*f*x + 2*f^2*x^2)) + (2*c*f*(-I + 2*f*x) + d*(-1 - (2*I)*f*x + 2*f^2*x^
2))*Tan[e + f*x])/(8*a*f^2*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.50, size = 50, normalized size = 0.60

method	result	size
risch	\(\frac {d \,x^{2}}{4 a}+\frac {c x}{2 a}+\frac {i \left (2 d x f +2 c f -i d \right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a \,f^{2}}\)	\(50\)
norman	\(\frac {\frac {d \,x^{2}}{4 a}+\frac {2 i c f +d}{4 a \,f^{2}}+\frac {d \,x^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{4 a}+\frac {\left (2 c f -i d \right ) \tan \left (f x +e \right )}{4 a \,f^{2}}+\frac {\left (2 c f +i d \right ) x}{4 a f}+\frac {d x \tan \left (f x +e \right )}{2 a f}+\frac {\left (2 c f -i d \right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{4 f a}}{1+\tan ^{2}\left (f x +e \right )}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a*d*x^2+1/2/a*c*x+1/8*I*(2*d*x*f+2*c*f-I*d)/a/f^2*exp(-2*I*(f*x+e))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.36, size = 57, normalized size = 0.68 \begin {gather*} \frac {{\left (2 i \, d f x + 2 i \, c f + 2 \, {\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*I*d*f*x + 2*I*c*f + 2*(d*f^2*x^2 + 2*c*f^2*x)*e^(2*I*f*x + 2*I*e) + d)*e^(-2*I*f*x - 2*I*e)/(a*f^2)

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Sympy [A]
time = 0.12, size = 92, normalized size = 1.10 \begin {gather*} \begin {cases} \frac {\left (2 i c f + 2 i d f x + d\right ) e^{- 2 i e} e^{- 2 i f x}}{8 a f^{2}} & \text {for}\: a f^{2} e^{2 i e} \neq 0 \\\frac {c x e^{- 2 i e}}{2 a} + \frac {d x^{2} e^{- 2 i e}}{4 a} & \text {otherwise} \end {cases} + \frac {c x}{2 a} + \frac {d x^{2}}{4 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*I*c*f + 2*I*d*f*x + d)*exp(-2*I*e)*exp(-2*I*f*x)/(8*a*f**2), Ne(a*f**2*exp(2*I*e), 0)), (c*x*exp
(-2*I*e)/(2*a) + d*x**2*exp(-2*I*e)/(4*a), True)) + c*x/(2*a) + d*x**2/(4*a)

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Giac [A]
time = 0.52, size = 65, normalized size = 0.77 \begin {gather*} \frac {{\left (2 \, d f^{2} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, c f^{2} x e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, d f x + 2 i \, c f + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*d*f^2*x^2*e^(2*I*f*x + 2*I*e) + 4*c*f^2*x*e^(2*I*f*x + 2*I*e) + 2*I*d*f*x + 2*I*c*f + d)*e^(-2*I*f*x -
2*I*e)/(a*f^2)

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Mupad [B]
time = 2.81, size = 105, normalized size = 1.25 \begin {gather*} \frac {d\,\cos \left (2\,e+2\,f\,x\right )+2\,d\,f^2\,x^2+2\,c\,f\,\sin \left (2\,e+2\,f\,x\right )+4\,c\,f^2\,x+2\,d\,f\,x\,\sin \left (2\,e+2\,f\,x\right )-d\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+c\,f\,\cos \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}+d\,f\,x\,\cos \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}}{8\,a\,f^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*cos(2*e + 2*f*x) - d*sin(2*e + 2*f*x)*1i + 2*d*f^2*x^2 + c*f*cos(2*e + 2*f*x)*2i + 2*c*f*sin(2*e + 2*f*x) +
4*c*f^2*x + d*f*x*cos(2*e + 2*f*x)*2i + 2*d*f*x*sin(2*e + 2*f*x))/(8*a*f^2)

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