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

Optimal. Leaf size=204 \[ \frac{8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}-\frac{B+i A}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-(I*A + B)/(5*f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((I/5)*A)/(c*f*(a + I*a*Tan[e + f
*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*A*Tan[e + f*x])/(15*a*c*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*
Tan[e + f*x])^(3/2)) + (8*A*Tan[e + f*x])/(15*a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.275143, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3588, 78, 45, 40, 39} \[ \frac{8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}-\frac{B+i A}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(I*A + B)/(5*f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((I/5)*A)/(c*f*(a + I*a*Tan[e + f
*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*A*Tan[e + f*x])/(15*a*c*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*
Tan[e + f*x])^(3/2)) + (8*A*Tan[e + f*x])/(15*a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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]

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 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 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(x*(a + b*x)^(m + 1)*(c + d*x)^(m +
1))/(2*a*c*(m + 1)), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; F
reeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{(a A) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(4 A) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(8 A) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=-\frac{i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac{i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 11.5343, size = 151, normalized size = 0.74 \[ \frac{\sec ^2(e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-150 A \sin (e+f x)-25 A \sin (3 (e+f x))-3 A \sin (5 (e+f x))+30 B \cos (e+f x)+15 B \cos (3 (e+f x))+3 B \cos (5 (e+f x)))}{240 a^2 c^3 f (\tan (e+f x)-i)^2 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(30*B*Cos[e + f*x] + 15*B*Cos[3*(e + f*x)] + 3*B*Cos[5
*(e + f*x)] - 150*A*Sin[e + f*x] - 25*A*Sin[3*(e + f*x)] - 3*A*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(
240*a^2*c^3*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.131, size = 124, normalized size = 0.6 \begin{align*}{\frac{8\,A \left ( \tan \left ( fx+e \right ) \right ) ^{7}+28\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}+35\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-3\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+15\,A\tan \left ( fx+e \right ) -3\,B}{15\,f{a}^{3}{c}^{3} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^3/c^3*(8*A*tan(f*x+e)^7+28*A*tan(f*x+e)^5+35*
A*tan(f*x+e)^3-3*B*tan(f*x+e)^2+15*A*tan(f*x+e)-3*B)/(-tan(f*x+e)+I)^4/(tan(f*x+e)+I)^4

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Maxima [B]  time = 2.64763, size = 446, normalized size = 2.19 \begin{align*} \frac{{\left (30 \,{\left (5 i \, A - B\right )} \cos \left (4 \, f x + 4 \, e\right ) + 5 \,{\left (5 i \, A - 3 \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) -{\left (150 \, A + 30 i \, B\right )} \sin \left (4 \, f x + 4 \, e\right ) -{\left (25 \, A + 15 i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) - 6 \, B\right )} \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 ) + 5 \,{\left (-5 i \, A - 3 \, B\right )} \cos \left (\frac{3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 30 \,{\left (-5 i \, A - B\right )} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left ({\left (150 \, A + 30 i \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) +{\left (25 \, A + 15 i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 30 \,{\left (5 i \, A - B\right )} \sin \left (4 \, f x + 4 \, e\right ) + 5 \,{\left (5 i \, A - 3 \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, A\right )} \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 ) +{\left (25 \, A - 15 i \, B\right )} \sin \left (\frac{3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (150 \, A - 30 i \, B\right )} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )}{480 \, a^{\frac{5}{2}} c^{\frac{5}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*((30*(5*I*A - B)*cos(4*f*x + 4*e) + 5*(5*I*A - 3*B)*cos(2*f*x + 2*e) - (150*A + 30*I*B)*sin(4*f*x + 4*e)
- (25*A + 15*I*B)*sin(2*f*x + 2*e) - 6*B)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 5*(-5*I*A -
3*B)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 30*(-5*I*A - B)*cos(1/2*arctan2(sin(2*f*x + 2*e),
cos(2*f*x + 2*e))) + ((150*A + 30*I*B)*cos(4*f*x + 4*e) + (25*A + 15*I*B)*cos(2*f*x + 2*e) + 30*(5*I*A - B)*si
n(4*f*x + 4*e) + 5*(5*I*A - 3*B)*sin(2*f*x + 2*e) + 6*A)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ (25*A - 15*I*B)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (150*A - 30*I*B)*sin(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e))))/(a^(5/2)*c^(5/2)*f)

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Fricas [A]  time = 1.50763, size = 540, normalized size = 2.65 \begin{align*} \frac{{\left ({\left (-3 i \, A - 3 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-28 i \, A - 18 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-175 i \, A - 45 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 96 \, B e^{\left (7 i \, f x + 7 i \, e\right )} - 60 \, B e^{\left (6 i \, f x + 6 i \, e\right )} + 96 \, B e^{\left (5 i \, f x + 5 i \, e\right )} +{\left (175 i \, A - 45 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (28 i \, A - 18 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\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}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*((-3*I*A - 3*B)*e^(12*I*f*x + 12*I*e) + (-28*I*A - 18*B)*e^(10*I*f*x + 10*I*e) + (-175*I*A - 45*B)*e^(8*
I*f*x + 8*I*e) + 96*B*e^(7*I*f*x + 7*I*e) - 60*B*e^(6*I*f*x + 6*I*e) + 96*B*e^(5*I*f*x + 5*I*e) + (175*I*A - 4
5*B)*e^(4*I*f*x + 4*I*e) + (28*I*A - 18*B)*e^(2*I*f*x + 2*I*e) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x - 5*I*e)/(a^3*c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (f x + e\right ) + A}{{\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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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