3.10.96 \(\int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [996]

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

[Out]

1/3*I*(c-I*c*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^(1/2)+1/3*I*(c-I*c*tan(f*x+e))^(1/2)/f/(a+I*a*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 {c-i c \tan (e+f x)}}{3 a f \sqrt {a+i a \tan (e+f x)}}+\frac {i \sqrt {c-i c \tan (e+f x)}}{3 f (a+i a \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.87, size = 68, normalized size = 0.76 \begin {gather*} \frac {(2+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 a f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.32, size = 74, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (3 i \tan \left (f x +e \right )-\left (\tan ^{2}\left (f x +e \right )\right )+2\right )}{3 f \,a^{2} \left (-\tan \left (f x +e \right )+i\right )^{3}}\)	\(74\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (3 i \tan \left (f x +e \right )-\left (\tan ^{2}\left (f x +e \right )\right )+2\right )}{3 f \,a^{2} \left (-\tan \left (f x +e \right )+i\right )^{3}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^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((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*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 = 1.34, size = 80, normalized size = 0.89 \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 (3 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{6 \, a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)/(a+I*a*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))*(3*I*e^(4*I*f*x + 4*I*e) + 4*I*e^(2*I*
f*x + 2*I*e) + I)*e^(-3*I*f*x - 3*I*e)/(a^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}{\left (i a \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((c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(3/2),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 6.31, size = 135, normalized size = 1.50 \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}}\,\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}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+4\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+3{}\mathrm {i}\right )}{12\,a^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^(1/2)/(a + a*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)*((c*(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)*4i + cos(4*e + 4*f*x)*1i + 4*sin(2*e + 2*
f*x) + sin(4*e + 4*f*x) + 3i))/(12*a^2*f)

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