3.10.63 \(\int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2} \, dx\) [963]

Optimal. Leaf size=62 \[ \frac {4 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 c f} \]

[Out]

4/3*I*a^2*(c-I*c*tan(f*x+e))^(3/2)/f-2/5*I*a^2*(c-I*c*tan(f*x+e))^(5/2)/c/f

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Rubi [A]
time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3603, 3568, 45} \begin {gather*} \frac {4 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(((4*I)/3)*a^2*(c - I*c*Tan[e + f*x])^(3/2))/f - (((2*I)/5)*a^2*(c - I*c*Tan[e + f*x])^(5/2))/(c*f)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2} \, dx &=\left (a^2 c^2\right ) \int \frac {\sec ^4(e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx\\ &=\frac {\left (i a^2\right ) \text {Subst}\left (\int (c-x) \sqrt {c+x} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {\left (i a^2\right ) \text {Subst}\left (\int \left (2 c \sqrt {c+x}-(c+x)^{3/2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {4 i a^2 (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 c f}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 80, normalized size = 1.29 \begin {gather*} -\frac {2 a^2 c \sec (e+f x) (\cos (e-f x)-i \sin (e-f x)) (-7 i+3 \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 f (\cos (f x)+i \sin (f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(-2*a^2*c*Sec[e + f*x]*(Cos[e - f*x] - I*Sin[e - f*x])*(-7*I + 3*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15
*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]
time = 0.30, size = 47, normalized size = 0.76

method	result	size
derivativedivides	\(-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\)	\(47\)
default	\(-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*I/f*a^2/c*(1/5*(c-I*c*tan(f*x+e))^(5/2)-2/3*c*(c-I*c*tan(f*x+e))^(3/2))

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Maxima [A]
time = 0.29, size = 48, normalized size = 0.77 \begin {gather*} -\frac {2 i \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} - 10 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c\right )}}{15 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-2/15*I*(3*(-I*c*tan(f*x + e) + c)^(5/2)*a^2 - 10*(-I*c*tan(f*x + e) + c)^(3/2)*a^2*c)/(c*f)

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Fricas [A]
time = 1.54, size = 75, normalized size = 1.21 \begin {gather*} -\frac {8 \, \sqrt {2} {\left (-5 i \, a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{2} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-8/15*sqrt(2)*(-5*I*a^2*c*e^(2*I*f*x + 2*I*e) - 2*I*a^2*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4
*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- c \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**2*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

-a**2*(Integral(-c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2,
x) + Integral(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-I*c*sqrt(-I*c*tan(e + f*x) + c)*ta
n(e + f*x)**3, x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^2*(-I*c*tan(f*x + e) + c)^(3/2), x)

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Mupad [B]
time = 6.79, size = 168, normalized size = 2.71 \begin {gather*} \frac {8\,a^2\,c\,\sqrt {\frac {2\,c}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+\frac {{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,7{}\mathrm {i}+{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,5{}\mathrm {i}+2{}\mathrm {i}\right )}{15\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )\,\left (2\,{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}+2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+\frac {{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}}{2}+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

(8*a^2*c*((2*c)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(exp(- e*2i - f*x*2i)*1i + (exp(- e*4i - f*x*4i)*1i)/2 + 1i/2)
*(exp(e*2i + f*x*2i)*7i + exp(e*4i + f*x*4i)*5i + 2i))/(15*f*(exp(e*2i + f*x*2i)*1i + 1i)*(2*exp(- e*2i - f*x*
2i) + 2*exp(e*2i + f*x*2i) + exp(- e*4i - f*x*4i)/2 + exp(e*4i + f*x*4i)/2 + 3))

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