3.11.22 \(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx\) [1022]

Optimal. Leaf size=140 \[ \frac {i}{5 f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{5 a f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 \tan (e+f x)}{5 a^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3604, 47, 39} \begin {gather*} \frac {2 \tan (e+f x)}{5 a^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{5 a f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

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

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

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

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Mathematica [A]
time = 1.66, size = 99, normalized size = 0.71 \begin {gather*} \frac {(12-4 \cos (2 (e+f x))-3 i \sec (e+f x) \sin (3 (e+f x))+5 i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{20 a^2 c f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]

[Out]

((12 - 4*Cos[2*(e + f*x)] - (3*I)*Sec[e + f*x]*Sin[3*(e + f*x)] + (5*I)*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x
]])/(20*a^2*c*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.40, size = 118, normalized size = 0.84

method	result	size
derivativedivides	\(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \left (\tan ^{4}\left (f x +e \right )\right )-2 \left (\tan ^{5}\left (f x +e \right )\right )+6 i \left (\tan ^{2}\left (f x +e \right )\right )-\left (\tan ^{3}\left (f x +e \right )\right )+2 i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} c \left (-\tan \left (f x +e \right )+i\right )^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(118\)
default	\(-\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \left (\tan ^{4}\left (f x +e \right )\right )-2 \left (\tan ^{5}\left (f x +e \right )\right )+6 i \left (\tan ^{2}\left (f x +e \right )\right )-\left (\tan ^{3}\left (f x +e \right )\right )+2 i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} c \left (-\tan \left (f x +e \right )+i\right )^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.95, size = 131, normalized size = 0.94 \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 (-5 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 16 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 10 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 16 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 20 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 6 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{40 \, a^{3} c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-5*I*e^(8*I*f*x + 8*I*e) - 16*I*e^(7
*I*f*x + 7*I*e) + 10*I*e^(6*I*f*x + 6*I*e) - 16*I*e^(5*I*f*x + 5*I*e) + 20*I*e^(4*I*f*x + 4*I*e) + 6*I*e^(2*I*
f*x + 2*I*e) + I)*e^(-5*I*f*x - 5*I*e)/(a^3*c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(5/2)*sqrt(-I*c*tan(f*x + e) + c)), x)

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Mupad [B]
time = 5.96, size = 158, normalized size = 1.13 \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 )\,15{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,5{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+15\,\sin \left (2\,e+2\,f\,x\right )+5\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )-5{}\mathrm {i}\right )}{40\,a^3\,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(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(1/2)),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)*15i + cos(4
*e + 4*f*x)*5i + cos(6*e + 6*f*x)*1i + 15*sin(2*e + 2*f*x) + 5*sin(4*e + 4*f*x) + sin(6*e + 6*f*x) - 5i))/(40*
a^3*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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