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

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

[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 = 0.05, 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 = {3723, 3479, 8} \[ \frac {i (c+d x)}{2 f (a+i a \tan (e+f x))}+\frac {(c+d x)^2}{4 a d}+\frac {d}{4 f^2 (a+i a \tan (e+f x))}-\frac {i d x}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*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*Ta
n[e + f*x]))

Rule 8

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

Rule 3479

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 3723

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 = 0.44, size = 96, normalized size = 1.14 \[ \frac {\left (2 c f (2 f x-i)+d \left (2 f^2 x^2-2 i f x-1\right )\right ) \tan (e+f x)-i \left (2 c f (2 f x+i)+d \left (2 f^2 x^2+2 i f x+1\right )\right )}{8 a f^2 (\tan (e+f x)-i)} \]

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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fricas [A]  time = 0.52, size = 55, normalized size = 0.65 \[ \frac {{\left (2 i \, d f x + 2 i \, c f + 2 \, {\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f^{2}} \]

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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giac [A]  time = 0.70, size = 65, normalized size = 0.77 \[ \frac {{\left (2 \, d f^{2} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, c f^{2} x e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, d f x + 2 i \, c f + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f^{2}} \]

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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maple [B]  time = 0.56, size = 187, normalized size = 2.23 \[ \frac {-\frac {i d \left (-\frac {\left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{2}+\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{4}+\frac {f x}{4}+\frac {e}{4}\right )}{f}+\frac {i c \left (\cos ^{2}\left (f x +e \right )\right )}{2}-\frac {i d e \left (\cos ^{2}\left (f x +e \right )\right )}{2 f}+\frac {d \left (\left (f x +e \right ) \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}+c \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {d e \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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

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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sympy [A]  time = 0.24, size = 97, normalized size = 1.15 \[ \begin {cases} - \frac {\left (- 2 i c f - 2 i d f x - d\right ) e^{- 2 i e} e^{- 2 i f x}}{8 a f^{2}} & \text {for}\: 8 a f^{2} e^{2 i e} \neq 0 \\\frac {c x e^{- 2 i e}}{2 a} + \frac {d x^{2} e^{- 2 i e}}{4 a} & \text {otherwise} \end {cases} + \frac {c x}{2 a} + \frac {d x^{2}}{4 a} \]

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(8*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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