3.8.63 \(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) [763]

Optimal. Leaf size=142 \[ -\frac {8 a^3 (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {8 a^3 (i A+2 B)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^3 (i A+5 B)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B}{c^3 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

2*a^3*B/c^3/f/(c-I*c*tan(f*x+e))^(1/2)-8/7*a^3*(I*A+B)/f/(c-I*c*tan(f*x+e))^(7/2)+8/5*a^3*(I*A+2*B)/c/f/(c-I*c
*tan(f*x+e))^(5/2)-2/3*a^3*(I*A+5*B)/c^2/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.14, antiderivative size = 142, 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)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {8 a^3 (2 B+i A)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {8 a^3 (B+i A)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a^3 B}{c^3 f \sqrt {c-i c \tan (e+f x)}} \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])^(7/2),x]

[Out]

(-8*a^3*(I*A + B))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) + (8*a^3*(I*A + 2*B))/(5*c*f*(c - I*c*Tan[e + f*x])^(5/2
)) - (2*a^3*(I*A + 5*B))/(3*c^2*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a^3*B)/(c^3*f*Sqrt[c - I*c*Tan[e + f*x]])

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))^{7/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^2 (A+B x)}{(c-i c x)^{9/2}} \, 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)^{9/2}}-\frac {4 a^2 (A-2 i B)}{c (c-i c x)^{7/2}}+\frac {a^2 (A-5 i B)}{c^2 (c-i c x)^{5/2}}+\frac {i a^2 B}{c^3 (c-i c x)^{3/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {8 a^3 (i A+B)}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {8 a^3 (i A+2 B)}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^3 (i A+5 B)}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B}{c^3 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 6.20, size = 141, normalized size = 0.99 \begin {gather*} \frac {a^3 \cos (e+f x) (i (A+13 i B) \cos (e+f x)+(-23 i A+89 B) \cos (3 (e+f x))+14 (A-2 i B+(A-17 i B) \cos (2 (e+f x))) \sin (e+f x)) (\cos (4 e+7 f x)+i \sin (4 e+7 f x)) \sqrt {c-i c \tan (e+f x)}}{105 c^4 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])^(7/2),x]

[Out]

(a^3*Cos[e + f*x]*(I*(A + (13*I)*B)*Cos[e + f*x] + ((-23*I)*A + 89*B)*Cos[3*(e + f*x)] + 14*(A - (2*I)*B + (A
- (17*I)*B)*Cos[2*(e + f*x)])*Sin[e + f*x])*(Cos[4*e + 7*f*x] + I*Sin[4*e + 7*f*x])*Sqrt[c - I*c*Tan[e + f*x]]
)/(105*c^4*f*(Cos[f*x] + I*Sin[f*x])^3)

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Maple [A]
time = 0.26, size = 105, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (-\frac {4 c^{3} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {c \left (-5 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(105\)
default	\(\frac {2 i a^{3} \left (-\frac {4 c^{3} \left (-i B +A \right )}{7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}+\frac {4 c^{2} \left (-2 i B +A \right )}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {c \left (-5 i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(105\)
risch	\(-\frac {a^{3} \left (15 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+15 B \,{\mathrm e}^{6 i \left (f x +e \right )}+3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-39 B \,{\mathrm e}^{4 i \left (f x +e \right )}-4 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+52 B \,{\mathrm e}^{2 i \left (f x +e \right )}+8 i A -104 B \right ) \sqrt {2}}{210 c^{3} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(115\)

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))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2*I/f*a^3/c^3*(-4/7*c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(7/2)+4/5*c^2*(A-2*I*B)/(c-I*c*tan(f*x+e))^(5/2)-1/3*c*(A-5
*I*B)/(c-I*c*tan(f*x+e))^(3/2)-I*B/(c-I*c*tan(f*x+e))^(1/2))

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Maxima [A]
time = 0.29, size = 106, normalized size = 0.75 \begin {gather*} -\frac {2 i \, {\left (105 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} B a^{3} + 35 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (A - 5 i \, B\right )} a^{3} c - 84 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 2 i \, B\right )} a^{3} c^{2} + 60 \, {\left (A - i \, B\right )} a^{3} c^{3}\right )}}{105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} 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))^(7/2),x, algorithm="maxima")

[Out]

-2/105*I*(105*I*(-I*c*tan(f*x + e) + c)^3*B*a^3 + 35*(-I*c*tan(f*x + e) + c)^2*(A - 5*I*B)*a^3*c - 84*(-I*c*ta
n(f*x + e) + c)*(A - 2*I*B)*a^3*c^2 + 60*(A - I*B)*a^3*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*c^3*f)

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Fricas [A]
time = 1.46, size = 128, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {2} {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} + 6 \, {\left (3 i \, A - 4 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (i \, A - 13 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A - 13 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, A - 13 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{210 \, c^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

-1/210*sqrt(2)*(15*(I*A + B)*a^3*e^(8*I*f*x + 8*I*e) + 6*(3*I*A - 4*B)*a^3*e^(6*I*f*x + 6*I*e) - (I*A - 13*B)*
a^3*e^(4*I*f*x + 4*I*e) + 4*(I*A - 13*B)*a^3*e^(2*I*f*x + 2*I*e) + 8*(I*A - 13*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*
I*e) + 1))/(c^4*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}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 A \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{3}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 B \tan ^{2}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{4}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i A \tan ^{2}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {i B \tan {\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i B \tan ^{3}{\left (e + f x \right )}}{i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 i c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{3} \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))**(7/2),x)

[Out]

-I*a**3*(Integral(I*A/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)
*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) +
Integral(-3*A*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x
) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x) + c))
, x) + Integral(A*tan(e + f*x)**3/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e
+ f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan(e + f*x)
+ c)), x) + Integral(-3*B*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*sqrt(-
I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I*c*tan
(e + f*x) + c)), x) + Integral(B*tan(e + f*x)**4/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*c**3*
sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c**3*sqrt(-I
*c*tan(e + f*x) + c)), x) + Integral(-3*I*A*tan(e + f*x)**2/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**
3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c
**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(I*B*tan(e + f*x)/(I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f
*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x
) + c**3*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-3*I*B*tan(e + f*x)**3/(I*c**3*sqrt(-I*c*tan(e + f*x) + c
)*tan(e + f*x)**3 - 3*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 3*I*c**3*sqrt(-I*c*tan(e + f*x) + c)*
tan(e + f*x) + c**3*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))^(7/2),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)^(7/2), x)

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Mupad [B]
time = 10.73, size = 161, normalized size = 1.13 \begin {gather*} -\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^3\,\left (A+B\,13{}\mathrm {i}\right )\,4{}\mathrm {i}}{105\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (3\,A+B\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{35\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (A+B\,13{}\mathrm {i}\right )\,2{}\mathrm {i}}{105\,c^4\,f}+\frac {a^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{14\,c^4\,f}-\frac {a^3\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (A+B\,13{}\mathrm {i}\right )\,1{}\mathrm {i}}{210\,c^4\,f}\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)^3)/(c - c*tan(e + f*x)*1i)^(7/2),x)

[Out]

-(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^3*(A + B*13i)*4i)/(105*c^4*f) + (a^3*exp(e*6i + f*x*6i)*(3*A
+ B*4i)*1i)/(35*c^4*f) + (a^3*exp(e*2i + f*x*2i)*(A + B*13i)*2i)/(105*c^4*f) + (a^3*exp(e*8i + f*x*8i)*(A - B
*1i)*1i)/(14*c^4*f) - (a^3*exp(e*4i + f*x*4i)*(A + B*13i)*1i)/(210*c^4*f))

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