3.11.28 \(\int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [1028]

Optimal. Leaf size=90 \[ -\frac {i \sqrt {a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 47, 37} \begin {gather*} -\frac {i \sqrt {a+i a \tan (e+f x)}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(3/2)) - ((I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(c
*f*Sqrt[c - I*c*Tan[e + f*x]])

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]

Rule 47

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 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 1.19, size = 89, normalized size = 0.99 \begin {gather*} \frac {(2+2 \cos (2 (e+f x))-i \sin (2 (e+f x))) (-i \cos (2 (e+f x))+\sin (2 (e+f x))) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 2*Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*Sqrt[a + I*a*Tan[e +
f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(6*c^2*f)

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Maple [A]
time = 0.36, size = 70, normalized size = 0.78

method	result	size
risch	\(-\frac {i \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+3\right )}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(64\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (3 i \tan \left (f x +e \right )+\tan ^{2}\left (f x +e \right )-2\right )}{3 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(70\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (3 i \tan \left (f x +e \right )+\tan ^{2}\left (f x +e \right )-2\right )}{3 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 0.93, size = 81, normalized size = 0.90 \begin {gather*} \frac {\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}} {\left (-i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 4 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, e^{\left (i \, f x + i \, e\right )}\right )}}{6 \, c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(5*I*f*x + 5*I*e) - 4*I*e^(3*I*f
*x + 3*I*e) - 3*I*e^(I*f*x + I*e))/(c^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*a*(tan(e + f*x) - I))/(-I*c*(tan(e + f*x) + I))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.05, size = 114, normalized size = 1.27 \begin {gather*} -\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+3{}\mathrm {i}\right )}{6\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*1i - sin(2
*e + 2*f*x) + 3i))/(6*c*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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