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

Optimal. Leaf size=102 \[ -\frac{(-2 B+i A) \sqrt{a+i a \tan (e+f x)}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.218365, antiderivative size = 102, normalized size of antiderivative = 1., 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} \[ -\frac{(-2 B+i A) \sqrt{a+i a \tan (e+f x)}}{3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{(B+i A) \sqrt{a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 5.5783, size = 101, normalized size = 0.99 \[ \frac{\cos (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) ((B-2 i A) \cos (e+f x)-(A+2 i B) \sin (e+f x))}{3 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.132, size = 100, normalized size = 1. \begin{align*}{\frac{2\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}+3\,iA\tan \left ( fx+e \right ) +A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-iB-3\,B\tan \left ( fx+e \right ) -2\,A}{3\,f{c}^{2} \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+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(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)/c^2*(2*I*B*tan(f*x+e)^2+3*I*A*tan(f*x+e)+A*tan(f
*x+e)^2-I*B-3*B*tan(f*x+e)-2*A)/(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+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.39564, size = 244, normalized size = 2.39 \begin{align*} \frac{{\left ({\left (-i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-4 i \, A + 2 \, 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 (i \, f x + i \, e\right )}}{6 \, c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt{i \, a \tan \left (f x + e\right ) + a}}{{\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+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),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)^(3/2), x)