3.849 \(\int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}-\frac{B+i A}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{B+4 i A}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{B+4 i A}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.293649, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 39} \[ \frac{2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}-\frac{B+i A}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{B+4 i A}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{B+4 i A}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

-(I*A + B)/(3*f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + ((4*I)*A + B)/(15*c*f*(a + I*a*Ta
n[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)*A + B)/(15*a*c*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c -
I*c*Tan[e + f*x]]) + (2*(4*A - I*B)*Tan[e + f*x])/(15*a^2*c*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 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))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^{7/2} (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))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(a (4 A-i B)) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{7/2} (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))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(4 A-i B) \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 )}{5 f}\\ &=-\frac{i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i A+B}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(2 (4 A-i B)) \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 f}\\ &=-\frac{i A+B}{3 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 i A+B}{15 c f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i A+B}{15 a c f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{2 (4 A-i B) \tan (e+f x)}{15 a^2 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 11.907, size = 133, normalized size = 0.62 \[ -\frac{i \sqrt{c-i c \tan (e+f x)} (20 (A-i B) \cos (2 (e+f x))+(A-4 i B) \cos (4 (e+f x))+40 i A \sin (2 (e+f x))+4 i A \sin (4 (e+f x))-45 A+10 B \sin (2 (e+f x))+B \sin (4 (e+f x)))}{120 a^2 c^2 f \sqrt{a+i a \tan (e+f x)}} \]

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])^(3/2)),x]

[Out]

((-I/120)*(-45*A + 20*(A - I*B)*Cos[2*(e + f*x)] + (A - (4*I)*B)*Cos[4*(e + f*x)] + (40*I)*A*Sin[2*(e + f*x)]
+ 10*B*Sin[2*(e + f*x)] + (4*I)*A*Sin[4*(e + f*x)] + B*Sin[4*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*c^2*
f*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.125, size = 199, normalized size = 0.9 \begin{align*}{\frac{-{\frac{i}{15}} \left ( 8\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{6}-2\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{5}+2\,B \left ( \tan \left ( fx+e \right ) \right ) ^{6}+20\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}+8\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}-5\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}+5\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}+15\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+20\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-3\,iB\tan \left ( fx+e \right ) +3\,iA+12\,A\tan \left ( fx+e \right ) -3\,B \right ) }{f{a}^{3}{c}^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{4} \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))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^3/c^2*(8*I*A*tan(f*x+e)^6-2*I*B*tan(f*x+e)
^5+2*B*tan(f*x+e)^6+20*I*A*tan(f*x+e)^4+8*A*tan(f*x+e)^5-5*I*B*tan(f*x+e)^3+5*B*tan(f*x+e)^4+15*I*A*tan(f*x+e)
^2+20*A*tan(f*x+e)^3-3*I*B*tan(f*x+e)+3*I*A+12*A*tan(f*x+e)-3*B)/(-tan(f*x+e)+I)^4/(tan(f*x+e)+I)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.48528, size = 531, normalized size = 2.46 \begin{align*} \frac{{\left ({\left (-5 i \, A - 5 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-65 i \, A - 35 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-48 i \, A + 48 \, B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} +{\left (30 i \, A - 30 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-48 i \, A + 48 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} +{\left (110 i \, A - 10 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (23 i \, A - 13 \, 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 )}}{240 \, a^{3} 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))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/240*((-5*I*A - 5*B)*e^(10*I*f*x + 10*I*e) + (-65*I*A - 35*B)*e^(8*I*f*x + 8*I*e) + (-48*I*A + 48*B)*e^(7*I*f
*x + 7*I*e) + (30*I*A - 30*B)*e^(6*I*f*x + 6*I*e) + (-48*I*A + 48*B)*e^(5*I*f*x + 5*I*e) + (110*I*A - 10*B)*e^
(4*I*f*x + 4*I*e) + (23*I*A - 13*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^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))**(5/2)/(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{5}{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))^(5/2)/(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)^(5/2)*(-I*c*tan(f*x + e) + c)^(3/2)), x)