3.8.62 \(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [762]

Optimal. Leaf size=140 \[ -\frac {8 a^3 (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 a^3 (i A+2 B)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^3 (i A+5 B)}{c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a^3 B \sqrt {c-i c \tan (e+f x)}}{c^3 f} \]

[Out]

-2*a^3*(I*A+5*B)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-2*a^3*B*(c-I*c*tan(f*x+e))^(1/2)/c^3/f-8/5*a^3*(I*A+B)/f/(c-I*
c*tan(f*x+e))^(5/2)+8/3*a^3*(I*A+2*B)/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.15, antiderivative size = 140, 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^3 (5 B+i A)}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {8 a^3 (2 B+i A)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 a^3 (B+i A)}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^3 B \sqrt {c-i c \tan (e+f x)}}{c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^3*(I*A + B))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (8*a^3*(I*A + 2*B))/(3*c*f*(c - I*c*Tan[e + f*x])^(3/2
)) - (2*a^3*(I*A + 5*B))/(c^2*f*Sqrt[c - I*c*Tan[e + f*x]]) - (2*a^3*B*Sqrt[c - I*c*Tan[e + f*x]])/(c^3*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))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^2 (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {4 a^2 (A-i B)}{(c-i c x)^{7/2}}-\frac {4 a^2 (A-2 i B)}{c (c-i c x)^{5/2}}+\frac {a^2 (A-5 i B)}{c^2 (c-i c x)^{3/2}}+\frac {i a^2 B}{c^3 \sqrt {c-i c x}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {8 a^3 (i A+B)}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 a^3 (i A+2 B)}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^3 (i A+5 B)}{c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a^3 B \sqrt {c-i c \tan (e+f x)}}{c^3 f}\\ \end {align*}

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Mathematica [A]
time = 4.49, size = 135, normalized size = 0.96 \begin {gather*} \frac {a^3 (3 (A-11 i B) \cos (e+f x)+(11 A-91 i B) \cos (3 (e+f x))-10 i (A-14 i B+(A-17 i B) \cos (2 (e+f x))) \sin (e+f x)) (-i \cos (3 (e+2 f x))+\sin (3 (e+2 f x))) \sqrt {c-i c \tan (e+f x)}}{15 c^3 f (\cos (f x)+i \sin (f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(3*(A - (11*I)*B)*Cos[e + f*x] + (11*A - (91*I)*B)*Cos[3*(e + f*x)] - (10*I)*(A - (14*I)*B + (A - (17*I)*
B)*Cos[2*(e + f*x)])*Sin[e + f*x])*((-I)*Cos[3*(e + 2*f*x)] + Sin[3*(e + 2*f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/
(15*c^3*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]
time = 0.24, size = 105, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (i B \sqrt {c -i c \tan \left (f x +e \right )}+\frac {4 c^{2} \left (-2 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {4 c^{3} \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {c \left (-5 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(105\)
default	\(\frac {2 i a^{3} \left (i B \sqrt {c -i c \tan \left (f x +e \right )}+\frac {4 c^{2} \left (-2 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {4 c^{3} \left (-i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {c \left (-5 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 111, normalized size = 0.79 \begin {gather*} -\frac {2 i \, {\left (-\frac {15 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a^{3}}{c^{2}} + \frac {15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (A - 5 i \, B\right )} a^{3} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 2 i \, B\right )} a^{3} c + 12 \, {\left (A - i \, B\right )} a^{3} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{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))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/15*I*(-15*I*sqrt(-I*c*tan(f*x + e) + c)*B*a^3/c^2 + (15*(-I*c*tan(f*x + e) + c)^2*(A - 5*I*B)*a^3 - 20*(-I*
c*tan(f*x + e) + c)*(A - 2*I*B)*a^3*c + 12*(A - I*B)*a^3*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*c))/(c*f)

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Fricas [A]
time = 1.64, size = 106, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {2} {\left (3 \, {\left (i \, A + B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (i \, A + 11 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A + 11 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, A + 11 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/15*sqrt(2)*(3*(I*A + B)*a^3*e^(6*I*f*x + 6*I*e) - (I*A + 11*B)*a^3*e^(4*I*f*x + 4*I*e) + 4*(I*A + 11*B)*a^3
*e^(2*I*f*x + 2*I*e) + 8*(I*A + 11*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \frac {i A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 A \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{3}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 B \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{4}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i A \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {i B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i B \tan ^{3}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \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))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

-I*a**3*(Integral(I*A/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c
)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-3*A*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f
*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c
)), x) + Integral(A*tan(e + f*x)**3/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*ta
n(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-3*B*tan(e + f*x)**2/(-c**2*sq
rt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c
*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)**4/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*
c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-3*I*A*tan(e
+ f*x)**2/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*
x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(I*B*tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(
e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Inte
gral(-3*I*B*tan(e + f*x)**3/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*c**2*sqrt(-I*c*tan(e + f*
x) + c)*tan(e + f*x) + c**2*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))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 10.28, size = 208, normalized size = 1.49 \begin {gather*} -\frac {a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,8{}\mathrm {i}+88\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+44\,B\,\cos \left (2\,e+2\,f\,x\right )-11\,B\,\cos \left (4\,e+4\,f\,x\right )+3\,B\,\cos \left (6\,e+6\,f\,x\right )-4\,A\,\sin \left (2\,e+2\,f\,x\right )+A\,\sin \left (4\,e+4\,f\,x\right )-3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,44{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,11{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{15\,c^3\,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)^3)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

-(a^3*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*8i + 88*B + A*cos(2*e
+ 2*f*x)*4i - A*cos(4*e + 4*f*x)*1i + A*cos(6*e + 6*f*x)*3i + 44*B*cos(2*e + 2*f*x) - 11*B*cos(4*e + 4*f*x) +
3*B*cos(6*e + 6*f*x) - 4*A*sin(2*e + 2*f*x) + A*sin(4*e + 4*f*x) - 3*A*sin(6*e + 6*f*x) + B*sin(2*e + 2*f*x)*
44i - B*sin(4*e + 4*f*x)*11i + B*sin(6*e + 6*f*x)*3i))/(15*c^3*f)

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