3.812 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {(-6 B+i A) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

-1/7*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(7/2)-1/35*(I*A-6*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f/(
c-I*c*tan(f*x+e))^(5/2)

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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} \[ -\frac {(-6 B+i A) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

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])^(7/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) - ((I*A - 6*B)*(a + I*a*Tan[e + f
*x])^(5/2))/(35*c*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 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))^{7/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (A+6 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 )}{7 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(i A-6 B) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 12.02, size = 121, normalized size = 1.19 \[ \frac {a^2 \cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (6 e+8 f x)+i \sin (6 e+8 f x)) ((B-6 i A) \cos (e+f x)-(A+6 i B) \sin (e+f x))}{35 c^4 f (\cos (f x)+i \sin (f x))^2} \]

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])^(7/2),x]

[Out]

(a^2*Cos[e + f*x]*(((-6*I)*A + B)*Cos[e + f*x] - (A + (6*I)*B)*Sin[e + f*x])*(Cos[6*e + 8*f*x] + I*Sin[6*e + 8
*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(35*c^4*f*(Cos[f*x] + I*Sin[f*x])^2)

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fricas [A]  time = 1.89, size = 103, normalized size = 1.01 \[ \frac {{\left ({\left (-5 i \, A - 5 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + {\left (-12 i \, A + 2 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-7 i \, A + 7 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{70 \, c^{4} f} \]

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))^(7/2),x, algorithm="fricas")

[Out]

1/70*((-5*I*A - 5*B)*a^2*e^(9*I*f*x + 9*I*e) + (-12*I*A + 2*B)*a^2*e^(7*I*f*x + 7*I*e) + (-7*I*A + 7*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^4*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

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))^(7/2),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)^(7/2), x)

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maple [A]  time = 0.50, size = 115, normalized size = 1.13 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (i A \left (\tan ^{2}\left (f x +e \right )\right )+5 i B \tan \left (f x +e \right )-6 B \left (\tan ^{2}\left (f x +e \right )\right )+6 i A -5 A \tan \left (f x +e \right )-B \right )}{35 f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{5}} \]

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))^(7/2),x)

[Out]

-1/35*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^4*(1+tan(f*x+e)^2)*(I*A*tan(f*x+e)^2+5
*I*B*tan(f*x+e)-6*B*tan(f*x+e)^2+6*I*A-5*A*tan(f*x+e)-B)/(tan(f*x+e)+I)^5

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maxima [B]  time = 1.03, size = 167, normalized size = 1.64 \[ -\frac {{\left (350 \, {\left (A - i \, B\right )} a^{2} \cos \left (9 \, f x + 9 \, e\right ) + 140 \, {\left (6 \, A + i \, B\right )} a^{2} \cos \left (7 \, f x + 7 \, e\right ) + 490 \, {\left (A + i \, B\right )} a^{2} \cos \left (5 \, f x + 5 \, e\right ) + {\left (350 i \, A + 350 \, B\right )} a^{2} \sin \left (9 \, f x + 9 \, e\right ) + {\left (840 i \, A - 140 \, B\right )} a^{2} \sin \left (7 \, f x + 7 \, e\right ) + {\left (490 i \, A - 490 \, B\right )} a^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (-4900 i \, c^{4} \cos \left (2 \, f x + 2 \, e\right ) + 4900 \, c^{4} \sin \left (2 \, f x + 2 \, e\right ) - 4900 i \, c^{4}\right )} f} \]

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))^(7/2),x, algorithm="maxima")

[Out]

-(350*(A - I*B)*a^2*cos(9*f*x + 9*e) + 140*(6*A + I*B)*a^2*cos(7*f*x + 7*e) + 490*(A + I*B)*a^2*cos(5*f*x + 5*
e) + (350*I*A + 350*B)*a^2*sin(9*f*x + 9*e) + (840*I*A - 140*B)*a^2*sin(7*f*x + 7*e) + (490*I*A - 490*B)*a^2*s
in(5*f*x + 5*e))*sqrt(a)*sqrt(c)/((-4900*I*c^4*cos(2*f*x + 2*e) + 4900*c^4*sin(2*f*x + 2*e) - 4900*I*c^4)*f)

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mupad [B]  time = 10.28, size = 192, normalized size = 1.88 \[ -\frac {a^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (4\,e+4\,f\,x\right )\,7{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,5{}\mathrm {i}-7\,B\,\cos \left (4\,e+4\,f\,x\right )+5\,B\,\cos \left (6\,e+6\,f\,x\right )-7\,A\,\sin \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (4\,e+4\,f\,x\right )\,7{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,5{}\mathrm {i}\right )}{70\,c^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(7/2),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)*7i +
A*cos(6*e + 6*f*x)*5i - 7*B*cos(4*e + 4*f*x) + 5*B*cos(6*e + 6*f*x) - 7*A*sin(4*e + 4*f*x) - 5*A*sin(6*e + 6*
f*x) - B*sin(4*e + 4*f*x)*7i + B*sin(6*e + 6*f*x)*5i))/(70*c^3*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))**(7/2),x)

[Out]

Timed out

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