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

Optimal. Leaf size=202 \[ \frac {d (c+d x) e^{-2 i e-2 i f x}}{4 a^2 f^2}+\frac {d (c+d x) e^{-4 i e-4 i f x}}{32 a^2 f^2}+\frac {i (c+d x)^2 e^{-2 i e-2 i f x}}{4 a^2 f}+\frac {i (c+d x)^2 e^{-4 i e-4 i f x}}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {i d^2 e^{-2 i e-2 i f x}}{8 a^2 f^3}-\frac {i d^2 e^{-4 i e-4 i f x}}{128 a^2 f^3} \]

[Out]

-1/8*I*d^2*exp(-2*I*e-2*I*f*x)/a^2/f^3-1/128*I*d^2*exp(-4*I*e-4*I*f*x)/a^2/f^3+1/4*d*exp(-2*I*e-2*I*f*x)*(d*x+
c)/a^2/f^2+1/32*d*exp(-4*I*e-4*I*f*x)*(d*x+c)/a^2/f^2+1/4*I*exp(-2*I*e-2*I*f*x)*(d*x+c)^2/a^2/f+1/16*I*exp(-4*
I*e-4*I*f*x)*(d*x+c)^2/a^2/f+1/12*(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.21, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3729, 2176, 2194} \[ \frac {d (c+d x) e^{-2 i e-2 i f x}}{4 a^2 f^2}+\frac {d (c+d x) e^{-4 i e-4 i f x}}{32 a^2 f^2}+\frac {i (c+d x)^2 e^{-2 i e-2 i f x}}{4 a^2 f}+\frac {i (c+d x)^2 e^{-4 i e-4 i f x}}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {i d^2 e^{-2 i e-2 i f x}}{8 a^2 f^3}-\frac {i d^2 e^{-4 i e-4 i f x}}{128 a^2 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*d^2*E^((-2*I)*e - (2*I)*f*x))/(a^2*f^3) - ((I/128)*d^2*E^((-4*I)*e - (4*I)*f*x))/(a^2*f^3) + (d*E^((-2
*I)*e - (2*I)*f*x)*(c + d*x))/(4*a^2*f^2) + (d*E^((-4*I)*e - (4*I)*f*x)*(c + d*x))/(32*a^2*f^2) + ((I/4)*E^((-
2*I)*e - (2*I)*f*x)*(c + d*x)^2)/(a^2*f) + ((I/16)*E^((-4*I)*e - (4*I)*f*x)*(c + d*x)^2)/(a^2*f) + (c + d*x)^3
/(12*a^2*d)

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 3729

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^2}{(a+i a \tan (e+f x))^2} \, dx &=\int \left (\frac {(c+d x)^2}{4 a^2}+\frac {e^{-2 i e-2 i f x} (c+d x)^2}{2 a^2}+\frac {e^{-4 i e-4 i f x} (c+d x)^2}{4 a^2}\right ) \, dx\\ &=\frac {(c+d x)^3}{12 a^2 d}+\frac {\int e^{-4 i e-4 i f x} (c+d x)^2 \, dx}{4 a^2}+\frac {\int e^{-2 i e-2 i f x} (c+d x)^2 \, dx}{2 a^2}\\ &=\frac {i e^{-2 i e-2 i f x} (c+d x)^2}{4 a^2 f}+\frac {i e^{-4 i e-4 i f x} (c+d x)^2}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {(i d) \int e^{-4 i e-4 i f x} (c+d x) \, dx}{8 a^2 f}-\frac {(i d) \int e^{-2 i e-2 i f x} (c+d x) \, dx}{2 a^2 f}\\ &=\frac {d e^{-2 i e-2 i f x} (c+d x)}{4 a^2 f^2}+\frac {d e^{-4 i e-4 i f x} (c+d x)}{32 a^2 f^2}+\frac {i e^{-2 i e-2 i f x} (c+d x)^2}{4 a^2 f}+\frac {i e^{-4 i e-4 i f x} (c+d x)^2}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {d^2 \int e^{-4 i e-4 i f x} \, dx}{32 a^2 f^2}-\frac {d^2 \int e^{-2 i e-2 i f x} \, dx}{4 a^2 f^2}\\ &=-\frac {i d^2 e^{-2 i e-2 i f x}}{8 a^2 f^3}-\frac {i d^2 e^{-4 i e-4 i f x}}{128 a^2 f^3}+\frac {d e^{-2 i e-2 i f x} (c+d x)}{4 a^2 f^2}+\frac {d e^{-4 i e-4 i f x} (c+d x)}{32 a^2 f^2}+\frac {i e^{-2 i e-2 i f x} (c+d x)^2}{4 a^2 f}+\frac {i e^{-4 i e-4 i f x} (c+d x)^2}{16 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 282, normalized size = 1.40 \[ \frac {\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac {2}{3} f^3 x \left (3 c^2+3 c d x+d^2 x^2\right ) (\cos (2 e)+i \sin (2 e))+\frac {1}{16} (\cos (2 e)-i \sin (2 e)) \cos (4 f x) ((2+2 i) c f+(2+2 i) d f x+d) ((2+2 i) c f+d ((2+2 i) f x-i))-\frac {1}{16} i (\cos (2 e)-i \sin (2 e)) \sin (4 f x) ((2+2 i) c f+(2+2 i) d f x+d) ((2+2 i) c f+d ((2+2 i) f x-i))-i \sin (2 f x) ((1+i) c f+(1+i) d f x+d) ((1+i) c f+d ((1+i) f x-i))+\cos (2 f x) ((1+i) c f+(1+i) d f x+d) ((1+i) c f+d ((1+i) f x-i))\right )}{8 f^3 (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((d + (1 + I)*c*f + (1 + I)*d*f*x)*((1 + I)*c*f + d*(-I + (1 + I)*f*
x))*Cos[2*f*x] + ((d + (2 + 2*I)*c*f + (2 + 2*I)*d*f*x)*((2 + 2*I)*c*f + d*(-I + (2 + 2*I)*f*x))*Cos[4*f*x]*(C
os[2*e] - I*Sin[2*e]))/16 + (2*f^3*x*(3*c^2 + 3*c*d*x + d^2*x^2)*(Cos[2*e] + I*Sin[2*e]))/3 - I*(d + (1 + I)*c
*f + (1 + I)*d*f*x)*((1 + I)*c*f + d*(-I + (1 + I)*f*x))*Sin[2*f*x] - (I/16)*(d + (2 + 2*I)*c*f + (2 + 2*I)*d*
f*x)*((2 + 2*I)*c*f + d*(-I + (2 + 2*I)*f*x))*(Cos[2*e] - I*Sin[2*e])*Sin[4*f*x]))/(8*f^3*(a + I*a*Tan[e + f*x
])^2)

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fricas [A]  time = 0.64, size = 160, normalized size = 0.79 \[ \frac {{\left (24 i \, d^{2} f^{2} x^{2} + 24 i \, c^{2} f^{2} + 12 \, c d f - 3 i \, d^{2} + {\left (48 i \, c d f^{2} + 12 \, d^{2} f\right )} x + 32 \, {\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (96 i \, d^{2} f^{2} x^{2} + 96 i \, c^{2} f^{2} + 96 \, c d f - 48 i \, d^{2} + {\left (192 i \, c d f^{2} + 96 \, d^{2} f\right )} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(24*I*d^2*f^2*x^2 + 24*I*c^2*f^2 + 12*c*d*f - 3*I*d^2 + (48*I*c*d*f^2 + 12*d^2*f)*x + 32*(d^2*f^3*x^3 +
3*c*d*f^3*x^2 + 3*c^2*f^3*x)*e^(4*I*f*x + 4*I*e) + (96*I*d^2*f^2*x^2 + 96*I*c^2*f^2 + 96*c*d*f - 48*I*d^2 + (1
92*I*c*d*f^2 + 96*d^2*f)*x)*e^(2*I*f*x + 2*I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f^3)

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giac [A]  time = 1.05, size = 227, normalized size = 1.12 \[ \frac {{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 i \, f x + 4 i \, e\right )} + 96 i \, d^{2} f^{2} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, d^{2} f^{2} x^{2} + 192 i \, c d f^{2} x e^{\left (2 i \, f x + 2 i \, e\right )} + 48 i \, c d f^{2} x + 96 i \, c^{2} f^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 96 \, d^{2} f x e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, c^{2} f^{2} + 12 \, d^{2} f x + 96 \, c d f e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c d f - 48 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, d^{2}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(32*d^2*f^3*x^3*e^(4*I*f*x + 4*I*e) + 96*c*d*f^3*x^2*e^(4*I*f*x + 4*I*e) + 96*c^2*f^3*x*e^(4*I*f*x + 4*I
*e) + 96*I*d^2*f^2*x^2*e^(2*I*f*x + 2*I*e) + 24*I*d^2*f^2*x^2 + 192*I*c*d*f^2*x*e^(2*I*f*x + 2*I*e) + 48*I*c*d
*f^2*x + 96*I*c^2*f^2*e^(2*I*f*x + 2*I*e) + 96*d^2*f*x*e^(2*I*f*x + 2*I*e) + 24*I*c^2*f^2 + 12*d^2*f*x + 96*c*
d*f*e^(2*I*f*x + 2*I*e) + 12*c*d*f - 48*I*d^2*e^(2*I*f*x + 2*I*e) - 3*I*d^2)*e^(-4*I*f*x - 4*I*e)/(a^2*f^3)

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maple [B]  time = 0.75, size = 968, normalized size = 4.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2/f*(-I/f*c*d*e*cos(f*x+e)^4+1/2*I*c^2*cos(f*x+e)^4-2*I/f^2*d^2*(-1/4*(f*x+e)^2*cos(f*x+e)^4+1/2*(f*x+e)*(
1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)-3/32*(f*x+e)^2+1/32*cos(f*x+e)^4+3/32*cos(f*x+e)^2
)+1/2*I/f^2*d^2*e^2*cos(f*x+e)^4-4*I/f*c*d*(-1/4*(f*x+e)*cos(f*x+e)^4+1/16*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f
*x+e)+3/32*f*x+3/32*e)+4*I/f^2*d^2*e*(-1/4*(f*x+e)*cos(f*x+e)^4+1/16*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+
3/32*f*x+3/32*e)+2/f^2*d^2*((f*x+e)^2*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)+1/8*(f*x+e)
*cos(f*x+e)^4-1/32*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)-15/64*f*x-15/64*e+3/8*(f*x+e)*cos(f*x+e)^2-3/16*si
n(f*x+e)*cos(f*x+e)-1/4*(f*x+e)^3)+4/f*c*d*((f*x+e)*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*
e)-3/16*(f*x+e)^2+1/16*cos(f*x+e)^4+3/16*cos(f*x+e)^2)-4/f^2*d^2*e*((f*x+e)*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))
*sin(f*x+e)+3/8*f*x+3/8*e)-3/16*(f*x+e)^2+1/16*cos(f*x+e)^4+3/16*cos(f*x+e)^2)+2*c^2*(1/4*(cos(f*x+e)^3+3/2*co
s(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)-4/f*c*d*e*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)+2/f
^2*d^2*e^2*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)-1/f^2*d^2*((f*x+e)^2*(1/2*sin(f*x+e)*c
os(f*x+e)+1/2*f*x+1/2*e)+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-1/3*(f*x+e)^3)-2/f*c
*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)+2/f^2*d^2*e*((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^2*(1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+
1/2*e)+2/f*c*d*e*(1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/f^2*d^2*e^2*(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)^2/(a+I*a*tan(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 = 2.99, size = 183, normalized size = 0.91 \[ \frac {c^2\,x}{4\,a^2}-{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}\,\left (\frac {\left (-8\,c^2\,f^2+c\,d\,f\,4{}\mathrm {i}+d^2\right )\,1{}\mathrm {i}}{128\,a^2\,f^3}-\frac {d^2\,x^2\,1{}\mathrm {i}}{16\,a^2\,f}+\frac {d\,x\,\left (-4\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^2\,f^2}\right )-{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,\left (\frac {\left (-2\,c^2\,f^2+c\,d\,f\,2{}\mathrm {i}+d^2\right )\,1{}\mathrm {i}}{8\,a^2\,f^3}-\frac {d^2\,x^2\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {d\,x\,\left (-2\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^2\,f^2}\right )+\frac {d^2\,x^3}{12\,a^2}+\frac {c\,d\,x^2}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*x)/(4*a^2) - exp(- e*4i - f*x*4i)*(((d^2 - 8*c^2*f^2 + c*d*f*4i)*1i)/(128*a^2*f^3) - (d^2*x^2*1i)/(16*a^2
*f) + (d*x*(d*1i - 4*c*f)*1i)/(32*a^2*f^2)) - exp(- e*2i - f*x*2i)*(((d^2 - 2*c^2*f^2 + c*d*f*2i)*1i)/(8*a^2*f
^3) - (d^2*x^2*1i)/(4*a^2*f) + (d*x*(d*1i - 2*c*f)*1i)/(4*a^2*f^2)) + (d^2*x^3)/(12*a^2) + (c*d*x^2)/(4*a^2)

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sympy [A]  time = 0.45, size = 420, normalized size = 2.08 \[ \begin {cases} \frac {\left (\left (64 i a^{2} c^{2} f^{5} e^{2 i e} + 128 i a^{2} c d f^{5} x e^{2 i e} + 32 a^{2} c d f^{4} e^{2 i e} + 64 i a^{2} d^{2} f^{5} x^{2} e^{2 i e} + 32 a^{2} d^{2} f^{4} x e^{2 i e} - 8 i a^{2} d^{2} f^{3} e^{2 i e}\right ) e^{- 4 i f x} + \left (256 i a^{2} c^{2} f^{5} e^{4 i e} + 512 i a^{2} c d f^{5} x e^{4 i e} + 256 a^{2} c d f^{4} e^{4 i e} + 256 i a^{2} d^{2} f^{5} x^{2} e^{4 i e} + 256 a^{2} d^{2} f^{4} x e^{4 i e} - 128 i a^{2} d^{2} f^{3} e^{4 i e}\right ) e^{- 2 i f x}\right ) e^{- 6 i e}}{1024 a^{4} f^{6}} & \text {for}\: 1024 a^{4} f^{6} e^{6 i e} \neq 0 \\\frac {x^{3} \left (2 d^{2} e^{2 i e} + d^{2}\right ) e^{- 4 i e}}{12 a^{2}} + \frac {x^{2} \left (2 c d e^{2 i e} + c d\right ) e^{- 4 i e}}{4 a^{2}} + \frac {x \left (2 c^{2} e^{2 i e} + c^{2}\right ) e^{- 4 i e}}{4 a^{2}} & \text {otherwise} \end {cases} + \frac {c^{2} x}{4 a^{2}} + \frac {c d x^{2}}{4 a^{2}} + \frac {d^{2} x^{3}}{12 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((64*I*a**2*c**2*f**5*exp(2*I*e) + 128*I*a**2*c*d*f**5*x*exp(2*I*e) + 32*a**2*c*d*f**4*exp(2*I*e) +
64*I*a**2*d**2*f**5*x**2*exp(2*I*e) + 32*a**2*d**2*f**4*x*exp(2*I*e) - 8*I*a**2*d**2*f**3*exp(2*I*e))*exp(-4*
I*f*x) + (256*I*a**2*c**2*f**5*exp(4*I*e) + 512*I*a**2*c*d*f**5*x*exp(4*I*e) + 256*a**2*c*d*f**4*exp(4*I*e) +
256*I*a**2*d**2*f**5*x**2*exp(4*I*e) + 256*a**2*d**2*f**4*x*exp(4*I*e) - 128*I*a**2*d**2*f**3*exp(4*I*e))*exp(
-2*I*f*x))*exp(-6*I*e)/(1024*a**4*f**6), Ne(1024*a**4*f**6*exp(6*I*e), 0)), (x**3*(2*d**2*exp(2*I*e) + d**2)*e
xp(-4*I*e)/(12*a**2) + x**2*(2*c*d*exp(2*I*e) + c*d)*exp(-4*I*e)/(4*a**2) + x*(2*c**2*exp(2*I*e) + c**2)*exp(-
4*I*e)/(4*a**2), True)) + c**2*x/(4*a**2) + c*d*x**2/(4*a**2) + d**2*x**3/(12*a**2)

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