3.11.40 \(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [1040]

Optimal. Leaf size=234 \[ \frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(I/3)/(f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((4*I)/3)/(a*f*Sqrt[a + I*a*Tan[e + f*x]
]*(c - I*c*Tan[e + f*x])^(5/2)) - (((4*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*f*(c - I*c*Tan[e + f*x])^(5/2))
- (((8*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((8*I)/15)*Sqrt[a + I*a*Ta
n[e + f*x]])/(a^2*c^2*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 {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {8 \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 2.90, size = 130, normalized size = 0.56 \begin {gather*} \frac {\sec (e+f x) (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-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 c^3 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])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(Sec[e + f*x]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(-45 + 20*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] - (40*I)*S
in[2*(e + f*x)] - (4*I)*Sin[4*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(120*a*c^3*f*(-I + Tan[e + f*x])*Sqrt[a
+ I*a*Tan[e + f*x]])

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Maple [A]
time = 0.41, size = 130, normalized size = 0.56

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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))^(3/2)/(c-I*c*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 = 1.36, size = 143, normalized size = 0.61 \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 (10 i \, f x + 10 i \, e\right )} - 23 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 110 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 30 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 65 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),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))*(-3*I*e^(10*I*f*x + 10*I*e) - 23*I*e
^(8*I*f*x + 8*I*e) - 110*I*e^(6*I*f*x + 6*I*e) + 48*I*e^(5*I*f*x + 5*I*e) - 30*I*e^(4*I*f*x + 4*I*e) + 48*I*e^
(3*I*f*x + 3*I*e) + 65*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-3*I*f*x - 3*I*e)/(a^2*c^3*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 {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 5.53, size = 140, normalized size = 0.60 \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 )\,20{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+40\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-45{}\mathrm {i}\right )}{120\,a^2\,c^2\,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)^(3/2)*(c - c*tan(e + f*x)*1i)^(5/2)),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)*20i + cos(4
*e + 4*f*x)*1i + 40*sin(2*e + 2*f*x) + 4*sin(4*e + 4*f*x) - 45i))/(120*a^2*c^2*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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