3.825 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {(-8 B+i A) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

[Out]

-1/9*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(9/2)-1/63*(I*A-8*B)*(a+I*a*tan(f*x+e))^(7/2)/c/f/(
c-I*c*tan(f*x+e))^(7/2)

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Rubi [A]  time = 0.23, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3588, 78, 37} \[ -\frac {(-8 B+i A) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(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])^(7/2))/(9*f*(c - I*c*Tan[e + f*x])^(9/2)) - ((I*A - 8*B)*(a + I*a*Tan[e + f
*x])^(7/2))/(63*c*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 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))^{7/2} (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)^{5/2} (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))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac {(a (A+8 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(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))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac {(i A-8 B) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 12.67, size = 121, normalized size = 1.19 \[ \frac {a^3 \cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (8 e+11 f x)+i \sin (8 e+11 f x)) ((B-8 i A) \cos (e+f x)-(A+8 i B) \sin (e+f x))}{63 c^5 f (\cos (f x)+i \sin (f x))^3} \]

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])^(9/2),x]

[Out]

(a^3*Cos[e + f*x]*(((-8*I)*A + B)*Cos[e + f*x] - (A + (8*I)*B)*Sin[e + f*x])*(Cos[8*e + 11*f*x] + I*Sin[8*e +
11*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(63*c^5*f*(Cos[f*x] + I*Sin[f*x])^3)

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fricas [A]  time = 0.59, size = 103, normalized size = 1.01 \[ \frac {{\left ({\left (-7 i \, A - 7 \, B\right )} a^{3} e^{\left (11 i \, f x + 11 i \, e\right )} + {\left (-16 i \, A + 2 \, B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-9 i \, A + 9 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{126 \, c^{5} f} \]

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))^(9/2),x, algorithm="fricas")

[Out]

1/126*((-7*I*A - 7*B)*a^3*e^(11*I*f*x + 11*I*e) + (-16*I*A + 2*B)*a^3*e^(9*I*f*x + 9*I*e) + (-9*I*A + 9*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^5*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(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)^(7/2)/(-I*c*tan(f*x + e) + c)^(9/2), x)

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maple [A]  time = 0.51, size = 134, normalized size = 1.31 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (8 i B \left (\tan ^{3}\left (f x +e \right )\right )+6 i A \left (\tan ^{2}\left (f x +e \right )\right )+A \left (\tan ^{3}\left (f x +e \right )\right )-6 i B \tan \left (f x +e \right )+15 B \left (\tan ^{2}\left (f x +e \right )\right )-8 i A +15 A \tan \left (f x +e \right )+B \right )}{63 f \,c^{5} \left (\tan \left (f x +e \right )+i\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(9/2),x)

[Out]

-1/63/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^5*(1+tan(f*x+e)^2)*(8*I*B*tan(f*x+e)^3+6
*I*A*tan(f*x+e)^2+A*tan(f*x+e)^3-6*I*B*tan(f*x+e)+15*B*tan(f*x+e)^2-8*I*A+15*A*tan(f*x+e)+B)/(tan(f*x+e)+I)^6

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maxima [B]  time = 1.22, size = 167, normalized size = 1.64 \[ -\frac {{\left (882 \, {\left (A - i \, B\right )} a^{3} \cos \left (11 \, f x + 11 \, e\right ) + 252 \, {\left (8 \, A + i \, B\right )} a^{3} \cos \left (9 \, f x + 9 \, e\right ) + 1134 \, {\left (A + i \, B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) + {\left (882 i \, A + 882 \, B\right )} a^{3} \sin \left (11 \, f x + 11 \, e\right ) + {\left (2016 i \, A - 252 \, B\right )} a^{3} \sin \left (9 \, f x + 9 \, e\right ) + {\left (1134 i \, A - 1134 \, B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (-15876 i \, c^{5} \cos \left (2 \, f x + 2 \, e\right ) + 15876 \, c^{5} \sin \left (2 \, f x + 2 \, e\right ) - 15876 i \, c^{5}\right )} f} \]

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))^(9/2),x, algorithm="maxima")

[Out]

-(882*(A - I*B)*a^3*cos(11*f*x + 11*e) + 252*(8*A + I*B)*a^3*cos(9*f*x + 9*e) + 1134*(A + I*B)*a^3*cos(7*f*x +
7*e) + (882*I*A + 882*B)*a^3*sin(11*f*x + 11*e) + (2016*I*A - 252*B)*a^3*sin(9*f*x + 9*e) + (1134*I*A - 1134*
B)*a^3*sin(7*f*x + 7*e))*sqrt(a)*sqrt(c)/((-15876*I*c^5*cos(2*f*x + 2*e) + 15876*c^5*sin(2*f*x + 2*e) - 15876*
I*c^5)*f)

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mupad [B]  time = 11.48, size = 192, normalized size = 1.88 \[ -\frac {a^3\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}+A\,\cos \left (8\,e+8\,f\,x\right )\,7{}\mathrm {i}-9\,B\,\cos \left (6\,e+6\,f\,x\right )+7\,B\,\cos \left (8\,e+8\,f\,x\right )-9\,A\,\sin \left (6\,e+6\,f\,x\right )-7\,A\,\sin \left (8\,e+8\,f\,x\right )-B\,\sin \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}+B\,\sin \left (8\,e+8\,f\,x\right )\,7{}\mathrm {i}\right )}{126\,c^4\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(7/2))/(c - c*tan(e + f*x)*1i)^(9/2),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)*9i +
A*cos(8*e + 8*f*x)*7i - 9*B*cos(6*e + 6*f*x) + 7*B*cos(8*e + 8*f*x) - 9*A*sin(6*e + 6*f*x) - 7*A*sin(8*e + 8*
f*x) - B*sin(6*e + 6*f*x)*9i + B*sin(8*e + 8*f*x)*7i))/(126*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))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(9/2),x)

[Out]

Timed out

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