3.24 \(\int \frac {c+d x}{(a+i a \cot (e+f x))^2} \, dx\)

Optimal. Leaf size=151 \[ -\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {x (c+d x)}{4 a^2}+\frac {3 d}{16 f^2 \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {3 i d x}{16 a^2 f}-\frac {d x^2}{8 a^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}+\frac {d}{16 f^2 (a+i a \cot (e+f x))^2} \]

[Out]

3/16*I*d*x/a^2/f-1/8*d*x^2/a^2+1/4*x*(d*x+c)/a^2+1/16*d/f^2/(a+I*a*cot(f*x+e))^2-1/4*I*(d*x+c)/f/(a+I*a*cot(f*
x+e))^2+3/16*d/f^2/(a^2+I*a^2*cot(f*x+e))-1/4*I*(d*x+c)/f/(a^2+I*a^2*cot(f*x+e))

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Rubi [A]  time = 0.14, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3479, 8, 3730} \[ -\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {x (c+d x)}{4 a^2}+\frac {3 d}{16 f^2 \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {3 i d x}{16 a^2 f}-\frac {d x^2}{8 a^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}+\frac {d}{16 f^2 (a+i a \cot (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/16)*d*x)/(a^2*f) - (d*x^2)/(8*a^2) + (x*(c + d*x))/(4*a^2) + d/(16*f^2*(a + I*a*Cot[e + f*x])^2) - ((I
/4)*(c + d*x))/(f*(a + I*a*Cot[e + f*x])^2) + (3*d)/(16*f^2*(a^2 + I*a^2*Cot[e + f*x])) - ((I/4)*(c + d*x))/(f
*(a^2 + I*a^2*Cot[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 3730

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{u = IntHide[(a
+ b*Tan[e + f*x])^n, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[Dist[(c + d*x)^(m - 1), u, x], x], x]] /; Fr
eeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0] && ILtQ[n, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {c+d x}{(a+i a \cot (e+f x))^2} \, dx &=\frac {x (c+d x)}{4 a^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}-\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}-d \int \left (\frac {x}{4 a^2}-\frac {i}{4 f (a+i a \cot (e+f x))^2}-\frac {i}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}\right ) \, dx\\ &=-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}-\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {(i d) \int \frac {1}{(a+i a \cot (e+f x))^2} \, dx}{4 f}+\frac {(i d) \int \frac {1}{a^2+i a^2 \cot (e+f x)} \, dx}{4 f}\\ &=-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}+\frac {d}{16 f^2 (a+i a \cot (e+f x))^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}+\frac {d}{8 f^2 \left (a^2+i a^2 \cot (e+f x)\right )}-\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {(i d) \int 1 \, dx}{8 a^2 f}+\frac {(i d) \int \frac {1}{a+i a \cot (e+f x)} \, dx}{8 a f}\\ &=\frac {i d x}{8 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}+\frac {d}{16 f^2 (a+i a \cot (e+f x))^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}+\frac {3 d}{16 f^2 \left (a^2+i a^2 \cot (e+f x)\right )}-\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}+\frac {(i d) \int 1 \, dx}{16 a^2 f}\\ &=\frac {3 i d x}{16 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}+\frac {d}{16 f^2 (a+i a \cot (e+f x))^2}-\frac {i (c+d x)}{4 f (a+i a \cot (e+f x))^2}+\frac {3 d}{16 f^2 \left (a^2+i a^2 \cot (e+f x)\right )}-\frac {i (c+d x)}{4 f \left (a^2+i a^2 \cot (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 165, normalized size = 1.09 \[ \frac {8 i (2 c f+d (2 f x+i)) \cos (2 (e+f x))+(-4 i c f-4 i d f x+d) \cos (4 (e+f x))-16 c f \sin (2 (e+f x))+4 c f \sin (4 (e+f x))+16 c e f+16 c f^2 x-8 d e^2-8 i d \sin (2 (e+f x))-16 d f x \sin (2 (e+f x))+i d \sin (4 (e+f x))+4 d f x \sin (4 (e+f x))+8 d f^2 x^2}{64 a^2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*d*e^2 + 16*c*e*f + 16*c*f^2*x + 8*d*f^2*x^2 + (8*I)*(2*c*f + d*(I + 2*f*x))*Cos[2*(e + f*x)] + (d - (4*I)*
c*f - (4*I)*d*f*x)*Cos[4*(e + f*x)] - (8*I)*d*Sin[2*(e + f*x)] - 16*c*f*Sin[2*(e + f*x)] - 16*d*f*x*Sin[2*(e +
f*x)] + I*d*Sin[4*(e + f*x)] + 4*c*f*Sin[4*(e + f*x)] + 4*d*f*x*Sin[4*(e + f*x)])/(64*a^2*f^2)

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fricas [A]  time = 0.76, size = 69, normalized size = 0.46 \[ \frac {8 \, d f^{2} x^{2} + 16 \, c f^{2} x + {\left (-4 i \, d f x - 4 i \, c f + d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (16 i \, d f x + 16 i \, c f - 8 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{64 \, a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(8*d*f^2*x^2 + 16*c*f^2*x + (-4*I*d*f*x - 4*I*c*f + d)*e^(4*I*f*x + 4*I*e) + (16*I*d*f*x + 16*I*c*f - 8*d
)*e^(2*I*f*x + 2*I*e))/(a^2*f^2)

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giac [A]  time = 1.73, size = 108, normalized size = 0.72 \[ \frac {8 \, d f^{2} x^{2} + 16 \, c f^{2} x - 4 i \, d f x e^{\left (4 i \, f x + 4 i \, e\right )} + 16 i \, d f x e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, c f e^{\left (4 i \, f x + 4 i \, e\right )} + 16 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + d e^{\left (4 i \, f x + 4 i \, e\right )} - 8 \, d e^{\left (2 i \, f x + 2 i \, e\right )}}{64 \, a^{2} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(8*d*f^2*x^2 + 16*c*f^2*x - 4*I*d*f*x*e^(4*I*f*x + 4*I*e) + 16*I*d*f*x*e^(2*I*f*x + 2*I*e) - 4*I*c*f*e^(4
*I*f*x + 4*I*e) + 16*I*c*f*e^(2*I*f*x + 2*I*e) + d*e^(4*I*f*x + 4*I*e) - 8*d*e^(2*I*f*x + 2*I*e))/(a^2*f^2)

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maple [B]  time = 1.66, size = 390, normalized size = 2.58 \[ -\frac {\frac {2 i d \left (\frac {\left (f x +e \right ) \left (\sin ^{4}\left (f x +e \right )\right )}{4}+\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{16}-\frac {3 f x}{32}-\frac {3 e}{32}\right )}{f}+\frac {i c \left (\sin ^{4}\left (f x +e \right )\right )}{2}-\frac {i d e \left (\sin ^{4}\left (f x +e \right )\right )}{2 f}+\frac {2 d \left (\left (f x +e \right ) \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{16}+\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{16}-\left (f x +e \right ) \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {\left (\sin ^{4}\left (f x +e \right )\right )}{16}\right )}{f}+2 c \left (-\frac {\sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )}{4}+\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{8}+\frac {f x}{8}+\frac {e}{8}\right )-\frac {2 e d \left (-\frac {\sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )}{4}+\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{8}+\frac {f x}{8}+\frac {e}{8}\right )}{f}-\frac {d \left (\left (f x +e \right ) \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {\left (f x +e \right )^{2}}{4}+\frac {\left (\sin ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {e d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}}{a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^2/f*(2*I/f*d*(1/4*(f*x+e)*sin(f*x+e)^4+1/16*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)-3/32*f*x-3/32*e)+1/2
*I*c*sin(f*x+e)^4-1/2*I/f*d*e*sin(f*x+e)^4+2/f*d*((f*x+e)*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/16*(f*x
+e)^2+1/16*sin(f*x+e)^2-(f*x+e)*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-1/16*sin(f*x+e)^
4)+2*c*(-1/4*sin(f*x+e)*cos(f*x+e)^3+1/8*sin(f*x+e)*cos(f*x+e)+1/8*f*x+1/8*e)-2*e/f*d*(-1/4*sin(f*x+e)*cos(f*x
+e)^3+1/8*sin(f*x+e)*cos(f*x+e)+1/8*f*x+1/8*e)-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)+e/f*d*(-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*cot(f*x+e))^2,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 = 0.34, size = 102, normalized size = 0.68 \[ {\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (\frac {\left (2\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,f^2}+\frac {d\,x\,1{}\mathrm {i}}{4\,a^2\,f}\right )-{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (\frac {\left (4\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{64\,a^2\,f^2}+\frac {d\,x\,1{}\mathrm {i}}{16\,a^2\,f}\right )+\frac {d\,x^2}{8\,a^2}+\frac {c\,x}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(e*2i + f*x*2i)*(((d*1i + 2*c*f)*1i)/(8*a^2*f^2) + (d*x*1i)/(4*a^2*f)) - exp(e*4i + f*x*4i)*(((d*1i + 4*c*f
)*1i)/(64*a^2*f^2) + (d*x*1i)/(16*a^2*f)) + (d*x^2)/(8*a^2) + (c*x)/(4*a^2)

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sympy [A]  time = 0.36, size = 214, normalized size = 1.42 \[ \begin {cases} \frac {\left (128 i a^{2} c f^{3} e^{2 i e} + 128 i a^{2} d f^{3} x e^{2 i e} - 64 a^{2} d f^{2} e^{2 i e}\right ) e^{2 i f x} + \left (- 32 i a^{2} c f^{3} e^{4 i e} - 32 i a^{2} d f^{3} x e^{4 i e} + 8 a^{2} d f^{2} e^{4 i e}\right ) e^{4 i f x}}{512 a^{4} f^{4}} & \text {for}\: 512 a^{4} f^{4} \neq 0 \\\frac {x^{2} \left (d e^{4 i e} - 2 d e^{2 i e}\right )}{8 a^{2}} + \frac {x \left (c e^{4 i e} - 2 c e^{2 i e}\right )}{4 a^{2}} & \text {otherwise} \end {cases} + \frac {c x}{4 a^{2}} + \frac {d x^{2}}{8 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((128*I*a**2*c*f**3*exp(2*I*e) + 128*I*a**2*d*f**3*x*exp(2*I*e) - 64*a**2*d*f**2*exp(2*I*e))*exp(2*
I*f*x) + (-32*I*a**2*c*f**3*exp(4*I*e) - 32*I*a**2*d*f**3*x*exp(4*I*e) + 8*a**2*d*f**2*exp(4*I*e))*exp(4*I*f*x
))/(512*a**4*f**4), Ne(512*a**4*f**4, 0)), (x**2*(d*exp(4*I*e) - 2*d*exp(2*I*e))/(8*a**2) + x*(c*exp(4*I*e) -
2*c*exp(2*I*e))/(4*a**2), True)) + c*x/(4*a**2) + d*x**2/(8*a**2)

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