3.813 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

[Out]

-1/9*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(9/2)-1/63*(2*I*A-7*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f
/(c-I*c*tan(f*x+e))^(7/2)-1/315*(2*I*A-7*B)*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(5/2)

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Rubi [A]  time = 0.26, 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 = {3588, 78, 45, 37} \[ -\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {(-7 B+2 i A) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

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])^(9/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(9*f*(c - I*c*Tan[e + f*x])^(9/2)) - (((2*I)*A - 7*B)*(a + I*a*Tan[e
+ f*x])^(5/2))/(63*c*f*(c - I*c*Tan[e + f*x])^(7/2)) - (((2*I)*A - 7*B)*(a + I*a*Tan[e + f*x])^(5/2))/(315*c^
2*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))^{9/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac {(a (2 A+7 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 )}{9 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (2 A+7 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 )}{63 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-7 B) (a+i a \tan (e+f x))^{5/2}}{315 c^2 f (c-i c \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 11.52, size = 135, normalized size = 0.87 \[ \frac {a^2 \cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (7 e+9 f x)+i \sin (7 e+9 f x)) (-7 (2 A+7 i B) \sin (2 (e+f x))+7 (2 B-7 i A) \cos (2 (e+f x))-45 i A)}{630 c^5 f (\cos (f x)+i \sin (f x))^2} \]

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])^(9/2),x]

[Out]

(a^2*Cos[e + f*x]*((-45*I)*A + 7*((-7*I)*A + 2*B)*Cos[2*(e + f*x)] - 7*(2*A + (7*I)*B)*Sin[2*(e + f*x)])*(Cos[
7*e + 9*f*x] + I*Sin[7*e + 9*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(630*c^5*f*(Cos[f*x]
+ I*Sin[f*x])^2)

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fricas [A]  time = 0.89, size = 123, normalized size = 0.79 \[ \frac {{\left ({\left (-35 i \, A - 35 \, B\right )} a^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + {\left (-125 i \, A - 35 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-153 i \, A + 63 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-63 i \, A + 63 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{1260 \, c^{5} f} \]

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))^(9/2),x, algorithm="fricas")

[Out]

1/1260*((-35*I*A - 35*B)*a^2*e^(11*I*f*x + 11*I*e) + (-125*I*A - 35*B)*a^2*e^(9*I*f*x + 9*I*e) + (-153*I*A + 6
3*B)*a^2*e^(7*I*f*x + 7*I*e) + (-63*I*A + 63*B)*a^2*e^(5*I*f*x + 5*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^5*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

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))^(9/2),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)^(9/2), x)

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maple [A]  time = 0.49, size = 138, normalized size = 0.89 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (-47 A -42 i B \left (\tan ^{2}\left (f x +e \right )\right )-12 A \left (\tan ^{2}\left (f x +e \right )\right )-33 i A \tan \left (f x +e \right )+2 i A \left (\tan ^{3}\left (f x +e \right )\right )-42 B \tan \left (f x +e \right )-7 i B -7 B \left (\tan ^{3}\left (f x +e \right )\right )\right )}{315 f \,c^{5} \left (\tan \left (f x +e \right )+i\right )^{6}} \]

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))^(9/2),x)

[Out]

-1/315*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^5*(1+tan(f*x+e)^2)*(-47*A-42*I*B*tan(
f*x+e)^2-12*A*tan(f*x+e)^2-33*I*A*tan(f*x+e)+2*I*A*tan(f*x+e)^3-42*B*tan(f*x+e)-7*I*B-7*B*tan(f*x+e)^3)/(tan(f
*x+e)+I)^6

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maxima [A]  time = 1.04, size = 198, normalized size = 1.28 \[ -\frac {{\left ({\left (35 i \, A + 35 \, B\right )} a^{2} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 90 i \, A a^{2} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (63 i \, A - 63 \, B\right )} a^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 35 \, {\left (A - i \, B\right )} a^{2} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 90 \, A a^{2} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 63 \, {\left (A + i \, B\right )} a^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{1260 \, c^{\frac {9}{2}} f} \]

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))^(9/2),x, algorithm="maxima")

[Out]

-1/1260*((35*I*A + 35*B)*a^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 90*I*A*a^2*cos(7/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (63*I*A - 63*B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)) - 35*(A - I*B)*a^2*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 90*A*a^2*sin(7/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))) - 63*(A + I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/
(c^(9/2)*f)

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mupad [B]  time = 11.00, size = 217, normalized size = 1.40 \[ -\frac {a^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,90{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}-63\,B\,\cos \left (4\,e+4\,f\,x\right )+35\,B\,\cos \left (8\,e+8\,f\,x\right )-63\,A\,\sin \left (4\,e+4\,f\,x\right )-90\,A\,\sin \left (6\,e+6\,f\,x\right )-35\,A\,\sin \left (8\,e+8\,f\,x\right )-B\,\sin \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,35{}\mathrm {i}\right )}{1260\,c^4\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(9/2),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)*63i
+ A*cos(6*e + 6*f*x)*90i + A*cos(8*e + 8*f*x)*35i - 63*B*cos(4*e + 4*f*x) + 35*B*cos(8*e + 8*f*x) - 63*A*sin(4
*e + 4*f*x) - 90*A*sin(6*e + 6*f*x) - 35*A*sin(8*e + 8*f*x) - B*sin(4*e + 4*f*x)*63i + B*sin(8*e + 8*f*x)*35i)
)/(1260*c^4*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))**(9/2),x)

[Out]

Timed out

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