3.8.45 \(\int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [745]

Optimal. Leaf size=60 \[ -\frac {2 a (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a B}{c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 60, 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}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}} \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))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a*B)/(c*f*Sqrt[c - I*c*Tan[e + f*x]])

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

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Mathematica [A]
time = 1.23, size = 98, normalized size = 1.63 \begin {gather*} \frac {2 a \cos (e+f x) (\cos (f x)-i \sin (f x)) ((-i A+2 B) \cos (e+f x)-3 i B \sin (e+f x)) (\cos (2 e+3 f x)+i \sin (2 e+3 f x)) \sqrt {c-i c \tan (e+f x)}}{3 c^2 f} \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*Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*(((-I)*A + 2*B)*Cos[e + f*x] - (3*I)*B*Sin[e + f*x])*(Cos[2*e + 3*f*
x] + I*Sin[2*e + 3*f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*c^2*f)

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Maple [A]
time = 0.23, size = 53, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {2 i a \left (-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {c \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(53\)
default	\(\frac {2 i a \left (-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {c \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(53\)
risch	\(-\frac {a \left (i A \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,{\mathrm e}^{2 i \left (f x +e \right )}+i A -5 B \right ) \sqrt {2}}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(62\)

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

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Maxima [A]
time = 0.29, size = 47, normalized size = 0.78 \begin {gather*} -\frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a + {\left (A - i \, B\right )} a c\right )}}{3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} 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]

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

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Fricas [A]
time = 2.37, size = 78, normalized size = 1.30 \begin {gather*} \frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (i \, A - 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 5 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} 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="fricas")

[Out]

1/6*sqrt(2)*((-I*A - B)*a*e^(4*I*f*x + 4*I*e) - 2*(I*A - 2*B)*a*e^(2*I*f*x + 2*I*e) + (-I*A + 5*B)*a)*sqrt(c/(
e^(2*I*f*x + 2*I*e) + 1))/(c^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- \frac {i A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {i B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\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/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integ
ral(A*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integ
ral(B*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + In
tegral(-I*B*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), 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 = 10.19, size = 170, normalized size = 2.83 \begin {gather*} -\frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (B\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )-4\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )-2\,A\,\sin \left (2\,e+2\,f\,x\right )-A\,\sin \left (4\,e+4\,f\,x\right )-5\,B+A\,1{}\mathrm {i}+A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,2{}\mathrm {i}-B\,\sin \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+A\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )\,1{}\mathrm {i}\right )}{6\,c^2\,f} \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]

-(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2)*(A*1i - 5*B + A*(2*cos(e + f*x)^2
- 1)*2i - 4*B*(2*cos(e + f*x)^2 - 1) - 2*A*sin(2*e + 2*f*x) - A*sin(4*e + 4*f*x) - B*sin(2*e + 2*f*x)*4i + B*
sin(4*e + 4*f*x)*1i + A*(2*cos(2*e + 2*f*x)^2 - 1)*1i + B*(2*cos(2*e + 2*f*x)^2 - 1)))/(6*c^2*f)

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