3.11.31 \(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx\) [1031]

Optimal. Leaf size=147 \[ \frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

8/15*tan(f*x+e)/a^2/c/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/5*I/f/(a+I*a*tan(f*x+e))^(5/2)/(c-
I*c*tan(f*x+e))^(3/2)+4/15*tan(f*x+e)/a/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.10, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3604, 47, 40, 39} \begin {gather*} \frac {8 \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(I/5)/(f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*Tan[e + f*x])/(15*a*f*(a + I*a*Tan[e
+ f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (8*Tan[e + f*x])/(15*a^2*c*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 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /;
FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 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} (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \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 )}{15 a f}\\ &=\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c 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 = 2.65, size = 93, normalized size = 0.63 \begin {gather*} -\frac {i (-45+20 \cos (2 (e+f x))+\cos (4 (e+f x))+40 i \sin (2 (e+f x))+4 i \sin (4 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{120 a^2 c^2 f \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)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

((-1/120*I)*(-45 + 20*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] + (40*I)*Sin[2*(e + f*x)] + (4*I)*Sin[4*(e + f*x)])*
Sqrt[c - I*c*Tan[e + f*x]])/(a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.38, size = 130, normalized size = 0.88

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{5}\left (f x +e \right )\right )-8 \left (\tan ^{6}\left (f x +e \right )\right )+20 i \left (\tan ^{3}\left (f x +e \right )\right )-20 \left (\tan ^{4}\left (f x +e \right )\right )+12 i \tan \left (f x +e \right )-15 \left (\tan ^{2}\left (f x +e \right )\right )-3\right )}{15 f \,a^{3} c^{2} \left (-\tan \left (f x +e \right )+i\right )^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(130\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{5}\left (f x +e \right )\right )-8 \left (\tan ^{6}\left (f x +e \right )\right )+20 i \left (\tan ^{3}\left (f x +e \right )\right )-20 \left (\tan ^{4}\left (f x +e \right )\right )+12 i \tan \left (f x +e \right )-15 \left (\tan ^{2}\left (f x +e \right )\right )-3\right )}{15 f \,a^{3} c^{2} \left (-\tan \left (f x +e \right )+i\right )^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^3/c^2*(8*I*tan(f*x+e)^5-8*tan(f*x+e)^6+20*I*t
an(f*x+e)^3-20*tan(f*x+e)^4+12*I*tan(f*x+e)-15*tan(f*x+e)^2-3)/(-tan(f*x+e)+I)^4/(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(1/(a+I*a*tan(f*x+e))^(5/2)/(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 = 1.06, size = 143, normalized size = 0.97 \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 (10 i \, f x + 10 i \, e\right )} - 65 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 48 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 30 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 110 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 23 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{240 \, a^{3} c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/240*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^(10*I*f*x + 10*I*e) - 65*I*e
^(8*I*f*x + 8*I*e) - 48*I*e^(7*I*f*x + 7*I*e) + 30*I*e^(6*I*f*x + 6*I*e) - 48*I*e^(5*I*f*x + 5*I*e) + 110*I*e^
(4*I*f*x + 4*I*e) + 23*I*e^(2*I*f*x + 2*I*e) + 3*I)*e^(-5*I*f*x - 5*I*e)/(a^3*c^2*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}} \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(1/(a+I*a*tan(f*x+e))**(5/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*(-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(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.91, size = 163, normalized size = 1.11 \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 )\,85{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,20{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+95\,\sin \left (2\,e+2\,f\,x\right )+20\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )-60{}\mathrm {i}\right )}{240\,a^3\,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(1/((a + a*tan(e + f*x)*1i)^(5/2)*(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)*85i + cos(4
*e + 4*f*x)*20i + cos(6*e + 6*f*x)*3i + 95*sin(2*e + 2*f*x) + 20*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x) - 60i))
/(240*a^3*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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