3.814 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac {2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac {(-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

[Out]

-1/11*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(11/2)-1/99*(3*I*A-8*B)*(a+I*a*tan(f*x+e))^(5/2)/c
/f/(c-I*c*tan(f*x+e))^(9/2)-2/693*(3*I*A-8*B)*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(7/2)-2/3465*(
3*I*A-8*B)*(a+I*a*tan(f*x+e))^(5/2)/c^3/f/(c-I*c*tan(f*x+e))^(5/2)

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Rubi [A]  time = 0.29, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac {2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac {(-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(11*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((3*I)*A - 8*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(69
3*c^2*f*(c - I*c*Tan[e + f*x])^(7/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3465*c^3*f*(c - I*c*
Tan[e + f*x])^(5/2))

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 45

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 78

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] || LtQ
[p, n]))))

Rule 3588

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))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2} (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))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}+\frac {(a (3 A+8 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(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))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}+\frac {(2 a (3 A+8 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(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))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}+\frac {(2 a (3 A+8 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{693 c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac {(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 14.11, size = 156, normalized size = 0.75 \[ \frac {a^2 \cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (8 e+10 f x)+i \sin (8 e+10 f x)) (-(3 A+8 i B) (55 \sin (e+f x)+63 \sin (3 (e+f x)))+55 (B-24 i A) \cos (e+f x)+63 (3 B-8 i A) \cos (3 (e+f x)))}{13860 c^6 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

(a^2*Cos[e + f*x]*(55*((-24*I)*A + B)*Cos[e + f*x] + 63*((-8*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (8*I)*B)*(5
5*Sin[e + f*x] + 63*Sin[3*(e + f*x)]))*(Cos[8*e + 10*f*x] + I*Sin[8*e + 10*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sq
rt[c - I*c*Tan[e + f*x]])/(13860*c^6*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [A]  time = 0.99, size = 143, normalized size = 0.69 \[ \frac {{\left ({\left (-315 i \, A - 315 \, B\right )} a^{2} e^{\left (13 i \, f x + 13 i \, e\right )} + {\left (-1470 i \, A - 700 \, B\right )} a^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + {\left (-2640 i \, A + 110 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-2178 i \, A + 1188 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-693 i \, A + 693 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{27720 \, c^{6} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

1/27720*((-315*I*A - 315*B)*a^2*e^(13*I*f*x + 13*I*e) + (-1470*I*A - 700*B)*a^2*e^(11*I*f*x + 11*I*e) + (-2640
*I*A + 110*B)*a^2*e^(9*I*f*x + 9*I*e) + (-2178*I*A + 1188*B)*a^2*e^(7*I*f*x + 7*I*e) + (-693*I*A + 693*B)*a^2*
e^(5*I*f*x + 5*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(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)^(5/2)/(-I*c*tan(f*x + e) + c)^(11/2), x)

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maple [A]  time = 0.42, size = 161, normalized size = 0.77 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (6 i A \left (\tan ^{4}\left (f x +e \right )\right )-112 i B \left (\tan ^{3}\left (f x +e \right )\right )-16 B \left (\tan ^{4}\left (f x +e \right )\right )-135 i A \left (\tan ^{2}\left (f x +e \right )\right )-42 A \left (\tan ^{3}\left (f x +e \right )\right )-427 i B \tan \left (f x +e \right )+360 B \left (\tan ^{2}\left (f x +e \right )\right )-456 i A +273 A \tan \left (f x +e \right )+61 B \right )}{3465 f \,c^{6} \left (\tan \left (f x +e \right )+i\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x)

[Out]

-1/3465*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^6*(1+tan(f*x+e)^2)*(6*I*A*tan(f*x+e)
^4-112*I*B*tan(f*x+e)^3-16*B*tan(f*x+e)^4-135*I*A*tan(f*x+e)^2-42*A*tan(f*x+e)^3-427*I*B*tan(f*x+e)+360*B*tan(
f*x+e)^2-456*I*A+273*A*tan(f*x+e)+61*B)/(tan(f*x+e)+I)^7

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maxima [A]  time = 1.22, size = 276, normalized size = 1.33 \[ -\frac {{\left ({\left (315 i \, A + 315 \, B\right )} a^{2} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (1155 i \, A + 385 \, B\right )} a^{2} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (1485 i \, A - 495 \, B\right )} a^{2} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (693 i \, A - 693 \, B\right )} a^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 315 \, {\left (A - i \, B\right )} a^{2} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 385 \, {\left (3 \, A - i \, B\right )} a^{2} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 495 \, {\left (3 \, A + i \, B\right )} a^{2} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 693 \, {\left (A + i \, B\right )} a^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{27720 \, c^{\frac {11}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

-1/27720*((315*I*A + 315*B)*a^2*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1155*I*A + 385*B)*a^2
*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1485*I*A - 495*B)*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))) + (693*I*A - 693*B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 315*(A -
I*B)*a^2*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 385*(3*A - I*B)*a^2*sin(9/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - 495*(3*A + I*B)*a^2*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 693*(
A + I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(11/2)*f)

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mupad [B]  time = 12.85, size = 292, normalized size = 1.40 \[ -\frac {a^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (4\,e+4\,f\,x\right )\,693{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,1485{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,1155{}\mathrm {i}+A\,\cos \left (10\,e+10\,f\,x\right )\,315{}\mathrm {i}-693\,B\,\cos \left (4\,e+4\,f\,x\right )-495\,B\,\cos \left (6\,e+6\,f\,x\right )+385\,B\,\cos \left (8\,e+8\,f\,x\right )+315\,B\,\cos \left (10\,e+10\,f\,x\right )-693\,A\,\sin \left (4\,e+4\,f\,x\right )-1485\,A\,\sin \left (6\,e+6\,f\,x\right )-1155\,A\,\sin \left (8\,e+8\,f\,x\right )-315\,A\,\sin \left (10\,e+10\,f\,x\right )-B\,\sin \left (4\,e+4\,f\,x\right )\,693{}\mathrm {i}-B\,\sin \left (6\,e+6\,f\,x\right )\,495{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,385{}\mathrm {i}+B\,\sin \left (10\,e+10\,f\,x\right )\,315{}\mathrm {i}\right )}{27720\,c^5\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(5/2))/(c - c*tan(e + f*x)*1i)^(11/2),x)

[Out]

-(a^2*((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(4*e + 4*f*x)*693i
+ A*cos(6*e + 6*f*x)*1485i + A*cos(8*e + 8*f*x)*1155i + A*cos(10*e + 10*f*x)*315i - 693*B*cos(4*e + 4*f*x) -
495*B*cos(6*e + 6*f*x) + 385*B*cos(8*e + 8*f*x) + 315*B*cos(10*e + 10*f*x) - 693*A*sin(4*e + 4*f*x) - 1485*A*s
in(6*e + 6*f*x) - 1155*A*sin(8*e + 8*f*x) - 315*A*sin(10*e + 10*f*x) - B*sin(4*e + 4*f*x)*693i - B*sin(6*e + 6
*f*x)*495i + B*sin(8*e + 8*f*x)*385i + B*sin(10*e + 10*f*x)*315i))/(27720*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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(11/2),x)

[Out]

Timed out

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