3.8.42 \(\int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\) [742]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 45} \begin {gather*} \frac {2 a (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a B (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])*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(2*a*(I*A + B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*f) - (2*a*B*(c - I*c*Tan[e + f*x])^(5/2))/(5*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 3669

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

Rubi steps

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

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Mathematica [A]
time = 1.10, size = 97, normalized size = 1.56 \begin {gather*} \frac {2 a c (\cos (e)-i \sin (e)) (\cos (f x)-i \sin (f x)) (5 i A+2 B+3 i B \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 f (A \cos (e+f x)+B \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*c*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*((5*I)*A + 2*B + (3*I)*B*Tan[e + f*x])*(A + B*Tan[e + f*x])
*Sqrt[c - I*c*Tan[e + f*x]])/(15*f*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]
time = 0.45, size = 55, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-i B c +A c \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\)	\(55\)
default	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-i B c +A c \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 50, normalized size = 0.81 \begin {gather*} \frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a 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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.98, size = 82, normalized size = 1.32 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (5 \, {\left (-i \, A - B\right )} a c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A + B\right )} a 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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-4/15*sqrt(2)*(5*(-I*A - B)*a*c*e^(2*I*f*x + 2*I*e) + (-5*I*A + B)*a*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*} i a \left (\int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- i B 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))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 11.95, size = 99, normalized size = 1.60 \begin {gather*} \frac {4\,a\,c\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,5{}\mathrm {i}-B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}+5\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}{15\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a*c*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(A*5i - B + A*exp(e*2i + f*x*2
i)*5i + 5*B*exp(e*2i + f*x*2i)))/(15*f*(exp(e*2i + f*x*2i) + 1)^2)

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