3.841 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=150 \[ -\frac{B+i A}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{2 A \tan (e+f x)}{3 a c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

-(I*A + B)/(3*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + ((I/3)*A)/(c*f*(a + I*a*Tan[e + f
*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) + (2*A*Tan[e + f*x])/(3*a*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*T
an[e + f*x]])

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Rubi [A]  time = 0.245398, antiderivative size = 150, normalized size of antiderivative = 1., 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, 39} \[ -\frac{B+i A}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{2 A \tan (e+f x)}{3 a c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

-(I*A + B)/(3*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + ((I/3)*A)/(c*f*(a + I*a*Tan[e + f
*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) + (2*A*Tan[e + f*x])/(3*a*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*T
an[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 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))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(a A) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(2 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 )}{3 f}\\ &=-\frac{i A+B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{2 A \tan (e+f x)}{3 a c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.5646, size = 120, normalized size = 0.8 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\sin (2 (e+f x))-i \cos (2 (e+f x))) (9 A \tan (e+f x)+A \sin (3 (e+f x)) \sec (e+f x)-2 B \cos (2 (e+f x))-2 B)}{12 a c^2 f (\tan (e+f x)-i) \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*(-2*B - 2*B*Cos[2*(e + f*x)] + A*Sec[e + f*x]*Sin[3*(e + f*x)] + 9
*A*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(12*a*c^2*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.125, size = 113, normalized size = 0.8 \begin{align*} -{\frac{2\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}+5\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+3\,A\tan \left ( fx+e \right ) -B}{3\,f{a}^{2}{c}^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}\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))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-1/3/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^2/c^2*(2*A*tan(f*x+e)^5+5*A*tan(f*x+e)^3-B*ta
n(f*x+e)^2+3*A*tan(f*x+e)-B)/(-tan(f*x+e)+I)^3/(tan(f*x+e)+I)^3

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Maxima [A]  time = 2.55065, size = 267, normalized size = 1.78 \begin{align*} \frac{{\left (3 \,{\left (3 i \, A - B\right )} \cos \left (2 \, f x + 2 \, e\right ) -{\left (9 \, A + 3 i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) - 2 \, 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 ) + 3 \,{\left (-3 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 (9 \, A + 3 i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 3 \,{\left (3 i \, A - B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 2 \, A\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 (9 \, A - 3 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 )}{24 \, a^{\frac{3}{2}} c^{\frac{3}{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))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

1/24*((3*(3*I*A - B)*cos(2*f*x + 2*e) - (9*A + 3*I*B)*sin(2*f*x + 2*e) - 2*B)*cos(3/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e))) + 3*(-3*I*A - B)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((9*A + 3*I*B)*co
s(2*f*x + 2*e) + 3*(3*I*A - B)*sin(2*f*x + 2*e) + 2*A)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
(9*A - 3*I*B)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))/(a^(3/2)*c^(3/2)*f)

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Fricas [A]  time = 1.38303, size = 409, normalized size = 2.73 \begin{align*} \frac{{\left ({\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-10 i \, A - 4 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 8 \, B e^{\left (5 i \, f x + 5 i \, e\right )} - 6 \, B e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, B e^{\left (3 i \, f x + 3 i \, e\right )} +{\left (10 i \, A - 4 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - 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 (-3 i \, f x - 3 i \, e\right )}}{24 \, a^{2} c^{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))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/24*((-I*A - B)*e^(8*I*f*x + 8*I*e) + (-10*I*A - 4*B)*e^(6*I*f*x + 6*I*e) + 8*B*e^(5*I*f*x + 5*I*e) - 6*B*e^(
4*I*f*x + 4*I*e) + 8*B*e^(3*I*f*x + 3*I*e) + (10*I*A - 4*B)*e^(2*I*f*x + 2*I*e) + I*A - B)*sqrt(a/(e^(2*I*f*x
+ 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-3*I*f*x - 3*I*e)/(a^2*c^2*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))**(3/2)/(c-I*c*tan(f*x+e))**(3/2),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{3}{2}}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{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))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^(3/2)*(-I*c*tan(f*x + e) + c)^(3/2)), x)