3.9.26 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx\) [826]

Optimal. Leaf size=155 \[ -\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

-1/11*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(11/2)-1/99*(2*I*A-9*B)*(a+I*a*tan(f*x+e))^(7/2)/c
/f/(c-I*c*tan(f*x+e))^(9/2)-1/693*(2*I*A-9*B)*(a+I*a*tan(f*x+e))^(7/2)/c^2/f/(c-I*c*tan(f*x+e))^(7/2)

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Rubi [A]
time = 0.17, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \begin {gather*} -\frac {(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac {(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

-1/11*((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(f*(c - I*c*Tan[e + f*x])^(11/2)) - (((2*I)*A - 9*B)*(a + I*a*T
an[e + f*x])^(7/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (((2*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(7/2))/(69
3*c^2*f*(c - I*c*Tan[e + f*x])^(7/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}+\frac {(a (2 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{11 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}+\frac {(a (2 A+9 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{99 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 7.62, size = 135, normalized size = 0.87 \begin {gather*} \frac {a^3 \cos (e+f x) (-77 i A+9 (-9 i A+2 B) \cos (2 (e+f x))-9 (2 A+9 i B) \sin (2 (e+f x))) (\cos (9 e+12 f x)+i \sin (9 e+12 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{1386 c^6 f (\cos (f x)+i \sin (f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

(a^3*Cos[e + f*x]*((-77*I)*A + 9*((-9*I)*A + 2*B)*Cos[2*(e + f*x)] - 9*(2*A + (9*I)*B)*Sin[2*(e + f*x)])*(Cos[
9*e + 12*f*x] + I*Sin[9*e + 12*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(1386*c^6*f*(Cos[f
*x] + I*Sin[f*x])^3)

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Maple [A]
time = 0.42, size = 161, normalized size = 1.04

method	result	size
risch	\(-\frac {a^{3} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (63 i A \,{\mathrm e}^{10 i \left (f x +e \right )}+63 B \,{\mathrm e}^{10 i \left (f x +e \right )}+154 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+99 i A \,{\mathrm e}^{6 i \left (f x +e \right )}-99 B \,{\mathrm e}^{6 i \left (f x +e \right )}\right )}{2772 c^{5} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(119\)
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i A \left (\tan ^{4}\left (f x +e \right )\right )-63 i B \left (\tan ^{3}\left (f x +e \right )\right )-9 B \left (\tan ^{4}\left (f x +e \right )\right )-45 i A \left (\tan ^{2}\left (f x +e \right )\right )-14 A \left (\tan ^{3}\left (f x +e \right )\right )+63 i B \tan \left (f x +e \right )-144 B \left (\tan ^{2}\left (f x +e \right )\right )+79 i A -140 A \tan \left (f x +e \right )-9 B \right )}{693 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}\)	\(161\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i A \left (\tan ^{4}\left (f x +e \right )\right )-63 i B \left (\tan ^{3}\left (f x +e \right )\right )-9 B \left (\tan ^{4}\left (f x +e \right )\right )-45 i A \left (\tan ^{2}\left (f x +e \right )\right )-14 A \left (\tan ^{3}\left (f x +e \right )\right )+63 i B \tan \left (f x +e \right )-144 B \left (\tan ^{2}\left (f x +e \right )\right )+79 i A -140 A \tan \left (f x +e \right )-9 B \right )}{693 f \,c^{6} \left (i+\tan \left (f x +e \right )\right )^{7}}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/693*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^6*(1+tan(f*x+e)^2)*(2*I*A*tan(f*x+e)^4-
63*I*B*tan(f*x+e)^3-9*B*tan(f*x+e)^4-45*I*A*tan(f*x+e)^2-14*A*tan(f*x+e)^3+63*I*B*tan(f*x+e)-144*B*tan(f*x+e)^
2+79*I*A-140*A*tan(f*x+e)-9*B)/(I+tan(f*x+e))^7

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Maxima [A]
time = 0.67, size = 210, normalized size = 1.35 \begin {gather*} \frac {{\left (63 \, {\left (-i \, A - B\right )} a^{3} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 154 i \, A a^{3} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 99 \, {\left (-i \, A + B\right )} a^{3} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 63 \, {\left (A - i \, B\right )} a^{3} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 154 \, A a^{3} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 99 \, {\left (A + i \, B\right )} a^{3} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{2772 \, c^{\frac {11}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

1/2772*(63*(-I*A - B)*a^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 154*I*A*a^3*cos(9/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 99*(-I*A + B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ 63*(A - I*B)*a^3*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 154*A*a^3*sin(9/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) + 99*(A + I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(
c^(11/2)*f)

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Fricas [A]
time = 3.14, size = 131, normalized size = 0.85 \begin {gather*} -\frac {{\left (63 \, {\left (i \, A + B\right )} a^{3} e^{\left (13 i \, f x + 13 i \, e\right )} + 7 \, {\left (31 i \, A + 9 \, B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + 11 \, {\left (23 i \, A - 9 \, B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + 99 \, {\left (i \, A - B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2772 \, c^{6} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/2772*(63*(I*A + B)*a^3*e^(13*I*f*x + 13*I*e) + 7*(31*I*A + 9*B)*a^3*e^(11*I*f*x + 11*I*e) + 11*(23*I*A - 9*
B)*a^3*e^(9*I*f*x + 9*I*e) + 99*(I*A - B)*a^3*e^(7*I*f*x + 7*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e
^(2*I*f*x + 2*I*e) + 1))/(c^6*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(11/2),x)

[Out]

Timed out

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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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(11/2), x)

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Mupad [B]
time = 11.75, size = 217, normalized size = 1.40 \begin {gather*} -\frac {a^3\,\sqrt {\frac {a\,\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\,\cos \left (6\,e+6\,f\,x\right )\,99{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,154{}\mathrm {i}+A\,\cos \left (10\,e+10\,f\,x\right )\,63{}\mathrm {i}-99\,B\,\cos \left (6\,e+6\,f\,x\right )+63\,B\,\cos \left (10\,e+10\,f\,x\right )-99\,A\,\sin \left (6\,e+6\,f\,x\right )-154\,A\,\sin \left (8\,e+8\,f\,x\right )-63\,A\,\sin \left (10\,e+10\,f\,x\right )-B\,\sin \left (6\,e+6\,f\,x\right )\,99{}\mathrm {i}+B\,\sin \left (10\,e+10\,f\,x\right )\,63{}\mathrm {i}\right )}{2772\,c^5\,f\,\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}}} \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)^(7/2))/(c - c*tan(e + f*x)*1i)^(11/2),x)

[Out]

-(a^3*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(6*e + 6*f*x)*99i
+ A*cos(8*e + 8*f*x)*154i + A*cos(10*e + 10*f*x)*63i - 99*B*cos(6*e + 6*f*x) + 63*B*cos(10*e + 10*f*x) - 99*A*
sin(6*e + 6*f*x) - 154*A*sin(8*e + 8*f*x) - 63*A*sin(10*e + 10*f*x) - B*sin(6*e + 6*f*x)*99i + B*sin(10*e + 10
*f*x)*63i))/(2772*c^5*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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