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

Optimal. Leaf size=213 \[ -\frac{2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{15 a c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{(-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+i A}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.275042, antiderivative size = 213, 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, 37} \[ -\frac{2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{15 a c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 (-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{15 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{(-2 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+i A}{f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 10.8351, size = 128, normalized size = 0.6 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) ((3 A+2 i B) (3 \sin (3 (e+f x))-5 \sin (e+f x))+5 (B-6 i A) \cos (e+f x)+(-9 B+6 i A) \cos (3 (e+f x)))}{60 c^3 f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(5*((-6*I)*A + B)*Cos[e + f*x] + ((6*I)*A - 9*B)*Cos[3*(e + f*x)] + (
3*A + (2*I)*B)*(-5*Sin[e + f*x] + 3*Sin[3*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(60*c^3*f*Sqrt[a + I*a*Tan[
e + f*x]])

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Maple [A]  time = 0.188, size = 184, normalized size = 0.9 \begin{align*}{\frac{4\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{5}+12\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}+6\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}+2\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-8\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}+18\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+3\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-2\,iB\tan \left ( fx+e \right ) -7\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+6\,iA-3\,A\tan \left ( fx+e \right ) +B}{15\,af{c}^{3} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{2} \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))^(1/2)/(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/c^3*(4*I*B*tan(f*x+e)^5+12*I*A*tan(f*x+e)^4+6
*A*tan(f*x+e)^5+2*I*B*tan(f*x+e)^3-8*B*tan(f*x+e)^4+18*I*A*tan(f*x+e)^2+3*A*tan(f*x+e)^3-2*I*B*tan(f*x+e)-7*B*
tan(f*x+e)^2+6*I*A-3*A*tan(f*x+e)+B)/(-tan(f*x+e)+I)^2/(tan(f*x+e)+I)^4

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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))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.41208, size = 447, normalized size = 2.1 \begin{align*} \frac{{\left ({\left (-3 i \, A - 3 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-18 i \, A - 8 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-60 i \, A + 10 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (48 i \, A + 8 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - 30 i \, A e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (48 i \, A + 8 \, B\right )} e^{\left (i \, f x + i \, e\right )} + 15 i \, A - 15 \, 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 (-i \, f x - i \, e\right )}}{120 \, a 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))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/120*((-3*I*A - 3*B)*e^(8*I*f*x + 8*I*e) + (-18*I*A - 8*B)*e^(6*I*f*x + 6*I*e) + (-60*I*A + 10*B)*e^(4*I*f*x
+ 4*I*e) + (48*I*A + 8*B)*e^(3*I*f*x + 3*I*e) - 30*I*A*e^(2*I*f*x + 2*I*e) + (48*I*A + 8*B)*e^(I*f*x + I*e) +
15*I*A - 15*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-I*f*x - I*e)/(a*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))**(1/2)/(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}{\sqrt{i \, a \tan \left (f x + e\right ) + a}{\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))^(1/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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