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

Optimal. Leaf size=104 \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{(2 B+i A) \sqrt{c-i c \tan (e+f x)}}{3 a f \sqrt{a+i a \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.214955, antiderivative size = 104, 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} \[ \frac{(-B+i A) \sqrt{c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac{(2 B+i A) \sqrt{c-i c \tan (e+f x)}}{3 a f \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 4.14039, size = 81, normalized size = 0.78 \[ \frac{\sqrt{c-i c \tan (e+f x)} ((2 B+i A) \tan (e+f x)+2 A-i B)}{3 a f (\tan (e+f x)-i) \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.114, size = 103, 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{a}^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{3}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.32497, size = 348, normalized size = 3.35 \begin{align*} \frac{{\left ({\left (-4 i \, A - 2 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} +{\left (3 i \, A + 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-4 i \, A - 2 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} +{\left (4 i \, A + 2 \, 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 )}}{6 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((-4*I*A - 2*B)*e^(5*I*f*x + 5*I*e) + (3*I*A + 3*B)*e^(4*I*f*x + 4*I*e) + (-4*I*A - 2*B)*e^(3*I*f*x + 3*I*
e) + (4*I*A + 2*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*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((c-I*c*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(a+I*a*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 \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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