3.8.52 \(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [752]

Optimal. Leaf size=101 \[ -\frac {4 a^2 (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a^2 (i A+3 B) \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{3/2}}{3 c^2 f} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3669, 78} \begin {gather*} -\frac {2 a^2 (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {4 a^2 (B+i A)}{f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^2 B (c-i c \tan (e+f x))^{3/2}}{3 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]
time = 1.44, size = 138, normalized size = 1.37 \begin {gather*} \frac {a^2 (9 A-15 i B+(9 A-13 i B) \cos (2 (e+f x))+(-3 i A-7 B) \sin (2 (e+f x))) (-i \cos (e+3 f x)+\sin (e+3 f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 c f (\cos (f x)+i \sin (f x))^2 (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])^2*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(a^2*(9*A - (15*I)*B + (9*A - (13*I)*B)*Cos[2*(e + f*x)] + ((-3*I)*A - 7*B)*Sin[2*(e + f*x)])*((-I)*Cos[e + 3*
f*x] + Sin[e + 3*f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*c*f*(Cos[f*x] + I*Sin[f*x])^2*(A*Co
s[e + f*x] + B*Sin[e + f*x]))

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Maple [A]
time = 0.26, size = 93, normalized size = 0.92

method	result	size
derivativedivides	\(-\frac {2 i a^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-3 i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}+\frac {2 c^{2} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(93\)
default	\(-\frac {2 i a^{2} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-3 i B c \sqrt {c -i c \tan \left (f x +e \right )}+A c \sqrt {c -i c \tan \left (f x +e \right )}+\frac {2 c^{2} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 84, normalized size = 0.83 \begin {gather*} -\frac {2 i \, {\left (\frac {6 \, {\left (A - i \, B\right )} a^{2} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} + \frac {i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B a^{2} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 3 i \, B\right )} a^{2} c}{c}\right )}}{3 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-2/3*I*(6*(A - I*B)*a^2*c/sqrt(-I*c*tan(f*x + e) + c) + (I*(-I*c*tan(f*x + e) + c)^(3/2)*B*a^2 + 3*sqrt(-I*c*t
an(f*x + e) + c)*(A - 3*I*B)*a^2*c)/c)/(c*f)

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Fricas [A]
time = 4.13, size = 97, normalized size = 0.96 \begin {gather*} -\frac {2 \, \sqrt {2} {\left (3 \, {\left (i \, A + B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, {\left (3 i \, A + 5 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (3 i \, A + 5 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(2)*(3*(I*A + B)*a^2*e^(4*I*f*x + 4*I*e) + 3*(3*I*A + 5*B)*a^2*e^(2*I*f*x + 2*I*e) + 2*(3*I*A + 5*B)*
a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(2*I*f*x + 2*I*e) + c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- \frac {A}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (e + f x \right )}}{\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))**2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

-a**2*(Integral(-A/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(A*tan(e + f*x)**2/sqrt(-I*c*tan(e + f*x) + c), x
) + Integral(-B*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(B*tan(e + f*x)**3/sqrt(-I*c*tan(e + f*
x) + c), x) + Integral(-2*I*A*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-2*I*B*tan(e + f*x)**2/s
qrt(-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))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 10.77, size = 176, normalized size = 1.74 \begin {gather*} -\frac {2\,\sqrt {2}\,a^2\,\sqrt {\frac {c}{\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}\,\left (A\,6{}\mathrm {i}+10\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,9{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+15\,B\,\cos \left (2\,e+2\,f\,x\right )+3\,B\,\cos \left (4\,e+4\,f\,x\right )-9\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{3\,c\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )} \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)^2)/(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

-(2*2^(1/2)*a^2*(c/(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))^(1/2)*(A*6i + 10*B + A*cos(2*e + 2*f*x)*9i +
A*cos(4*e + 4*f*x)*3i + 15*B*cos(2*e + 2*f*x) + 3*B*cos(4*e + 4*f*x) - 9*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*
f*x) + B*sin(2*e + 2*f*x)*15i + B*sin(4*e + 4*f*x)*3i))/(3*c*f*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))

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