3.1.19 \(\int \frac {(c+d x)^2}{a+i a \tan (e+f x)} \, dx\) [19]

Optimal. Leaf size=137 \[ -\frac {d^2 x}{4 a f^2}-\frac {i (c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {i d^2}{4 f^3 (a+i a \tan (e+f x))}+\frac {d (c+d x)}{2 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))} \]

[Out]

-1/4*d^2*x/a/f^2-1/4*I*(d*x+c)^2/a/f+1/6*(d*x+c)^3/a/d-1/4*I*d^2/f^3/(a+I*a*tan(f*x+e))+1/2*d*(d*x+c)/f^2/(a+I
*a*tan(f*x+e))+1/2*I*(d*x+c)^2/f/(a+I*a*tan(f*x+e))

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Rubi [A]
time = 0.09, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3804, 3560, 8} \begin {gather*} \frac {d (c+d x)}{2 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}-\frac {i (c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {i d^2}{4 f^3 (a+i a \tan (e+f x))}-\frac {d^2 x}{4 a f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(d^2*x)/(a*f^2) - ((I/4)*(c + d*x)^2)/(a*f) + (c + d*x)^3/(6*a*d) - ((I/4)*d^2)/(f^3*(a + I*a*Tan[e + f*x
])) + (d*(c + d*x))/(2*f^2*(a + I*a*Tan[e + f*x])) + ((I/2)*(c + d*x)^2)/(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)^2}{a+i a \tan (e+f x)} \, dx &=\frac {(c+d x)^3}{6 a d}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}-\frac {(i d) \int \frac {c+d x}{a+i a \tan (e+f x)} \, dx}{f}\\ &=-\frac {i (c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}+\frac {d (c+d x)}{2 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}-\frac {d^2 \int \frac {1}{a+i a \tan (e+f x)} \, dx}{2 f^2}\\ &=-\frac {i (c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {i d^2}{4 f^3 (a+i a \tan (e+f x))}+\frac {d (c+d x)}{2 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}-\frac {d^2 \int 1 \, dx}{4 a f^2}\\ &=-\frac {d^2 x}{4 a f^2}-\frac {i (c+d x)^2}{4 a f}+\frac {(c+d x)^3}{6 a d}-\frac {i d^2}{4 f^3 (a+i a \tan (e+f x))}+\frac {d (c+d x)}{2 f^2 (a+i a \tan (e+f x))}+\frac {i (c+d x)^2}{2 f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 178, normalized size = 1.30 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left ((d+(1+i) c f+(1+i) d f x) ((1+i) c f+d (-i+(1+i) f x)) \cos (2 f x) (\cos (e)-i \sin (e))+\frac {4}{3} f^3 x \left (3 c^2+3 c d x+d^2 x^2\right ) (\cos (e)+i \sin (e))-i (d+(1+i) c f+(1+i) d f x) ((1+i) c f+d (-i+(1+i) f x)) (\cos (e)-i \sin (e)) \sin (2 f x)\right )}{8 f^3 (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.56, size = 108, normalized size = 0.79

method	result	size
risch	\(\frac {d^{2} x^{3}}{6 a}+\frac {d c \,x^{2}}{2 a}+\frac {c^{2} x}{2 a}+\frac {c^{3}}{6 d a}+\frac {i \left (2 d^{2} x^{2} f^{2}+4 c d \,f^{2} x -2 i d^{2} f x +2 c^{2} f^{2}-2 i c d f -d^{2}\right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a \,f^{3}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*d^2/a*x^3+1/2*d/a*c*x^2+1/2/a*c^2*x+1/6/d/a*c^3+1/8*I*(2*d^2*x^2*f^2+4*c*d*f^2*x+2*c^2*f^2-2*I*d^2*f*x-d^2
-2*I*c*d*f)/a/f^3*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)^2/(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 = 0.35, size = 107, normalized size = 0.78 \begin {gather*} \frac {{\left (6 i \, d^{2} f^{2} x^{2} + 6 i \, c^{2} f^{2} + 6 \, c d f - 3 i \, d^{2} - 6 \, {\left (-2 i \, c d f^{2} - d^{2} f\right )} x + 4 \, {\left (d^{2} f^{3} x^{3} + 3 \, c d f^{3} x^{2} + 3 \, c^{2} f^{3} x\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{24 \, a f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*I*d^2*f^2*x^2 + 6*I*c^2*f^2 + 6*c*d*f - 3*I*d^2 - 6*(-2*I*c*d*f^2 - d^2*f)*x + 4*(d^2*f^3*x^3 + 3*c*d*
f^3*x^2 + 3*c^2*f^3*x)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f^3)

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Sympy [A]
time = 0.17, size = 165, normalized size = 1.20 \begin {gather*} \begin {cases} \frac {\left (2 i c^{2} f^{2} + 4 i c d f^{2} x + 2 c d f + 2 i d^{2} f^{2} x^{2} + 2 d^{2} f x - i d^{2}\right ) e^{- 2 i e} e^{- 2 i f x}}{8 a f^{3}} & \text {for}\: a f^{3} e^{2 i e} \neq 0 \\\frac {c^{2} x e^{- 2 i e}}{2 a} + \frac {c d x^{2} e^{- 2 i e}}{2 a} + \frac {d^{2} x^{3} e^{- 2 i e}}{6 a} & \text {otherwise} \end {cases} + \frac {c^{2} x}{2 a} + \frac {c d x^{2}}{2 a} + \frac {d^{2} x^{3}}{6 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*I*c**2*f**2 + 4*I*c*d*f**2*x + 2*c*d*f + 2*I*d**2*f**2*x**2 + 2*d**2*f*x - I*d**2)*exp(-2*I*e)*e
xp(-2*I*f*x)/(8*a*f**3), Ne(a*f**3*exp(2*I*e), 0)), (c**2*x*exp(-2*I*e)/(2*a) + c*d*x**2*exp(-2*I*e)/(2*a) + d
**2*x**3*exp(-2*I*e)/(6*a), True)) + c**2*x/(2*a) + c*d*x**2/(2*a) + d**2*x**3/(6*a)

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Giac [A]
time = 0.52, size = 123, normalized size = 0.90 \begin {gather*} \frac {{\left (4 \, d^{2} f^{3} x^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c d f^{3} x^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 12 \, c^{2} f^{3} x e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, d^{2} f^{2} x^{2} + 12 i \, c d f^{2} x + 6 i \, c^{2} f^{2} + 6 \, d^{2} f x + 6 \, c d f - 3 i \, d^{2}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{24 \, a f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(4*d^2*f^3*x^3*e^(2*I*f*x + 2*I*e) + 12*c*d*f^3*x^2*e^(2*I*f*x + 2*I*e) + 12*c^2*f^3*x*e^(2*I*f*x + 2*I*e
) + 6*I*d^2*f^2*x^2 + 12*I*c*d*f^2*x + 6*I*c^2*f^2 + 6*d^2*f*x + 6*c*d*f - 3*I*d^2)*e^(-2*I*f*x - 2*I*e)/(a*f^
3)

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Mupad [B]
time = 3.08, size = 241, normalized size = 1.76 \begin {gather*} \frac {12\,c^2\,f^3\,x-3\,d^2\,\sin \left (2\,e+2\,f\,x\right )+6\,c^2\,f^2\,\sin \left (2\,e+2\,f\,x\right )+4\,d^2\,f^3\,x^3+6\,c\,d\,f\,\cos \left (2\,e+2\,f\,x\right )+6\,d^2\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )+12\,c\,d\,f^3\,x^2+6\,d^2\,f\,x\,\cos \left (2\,e+2\,f\,x\right )+12\,c\,d\,f^2\,x\,\sin \left (2\,e+2\,f\,x\right )-d^2\,\cos \left (2\,e+2\,f\,x\right )\,3{}\mathrm {i}+c^2\,f^2\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-c\,d\,f\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+d^2\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}-d^2\,f\,x\,\sin \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+c\,d\,f^2\,x\,\cos \left (2\,e+2\,f\,x\right )\,12{}\mathrm {i}}{24\,a\,f^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*c^2*f^3*x - 3*d^2*sin(2*e + 2*f*x) - d^2*cos(2*e + 2*f*x)*3i + c^2*f^2*cos(2*e + 2*f*x)*6i + 6*c^2*f^2*sin
(2*e + 2*f*x) + 4*d^2*f^3*x^3 + 6*c*d*f*cos(2*e + 2*f*x) - c*d*f*sin(2*e + 2*f*x)*6i + d^2*f^2*x^2*cos(2*e + 2
*f*x)*6i + 6*d^2*f^2*x^2*sin(2*e + 2*f*x) + 12*c*d*f^3*x^2 + 6*d^2*f*x*cos(2*e + 2*f*x) - d^2*f*x*sin(2*e + 2*
f*x)*6i + c*d*f^2*x*cos(2*e + 2*f*x)*12i + 12*c*d*f^2*x*sin(2*e + 2*f*x))/(24*a*f^3)

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