3.1039 \(\int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ -\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {3 i \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {3 i \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

I/(f*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) - (((3*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*f*(c
- I*c*Tan[e + f*x])^(5/2)) - (((2*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((
(2*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a*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 45

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 3523

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}{\sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {(3 c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {3 i \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}+\frac {6 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {3 i \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 c f}\\ &=\frac {i}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {3 i \sqrt {a+i a \tan (e+f x)}}{5 a f (c-i c \tan (e+f x))^{5/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{5 a c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 4.21, size = 106, normalized size = 0.57 \[ \frac {\sqrt {c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-5 \sin (e+f x)+3 \sin (3 (e+f x))-10 i \cos (e+f x)+2 i \cos (3 (e+f x)))}{20 c^3 f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*((-10*I)*Cos[e + f*x] + (2*I)*Cos[3*(e + f*x)] - 5*Sin[e + f*x] + 3*S
in[3*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(20*c^3*f*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [A]  time = 0.46, size = 122, normalized size = 0.66 \[ \frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 6 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 20 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 16 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 10 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, e^{\left (i \, f x + i \, e\right )} + 5 i\right )} e^{\left (-i \, f x - i \, e\right )}}{40 \, a c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.32, size = 118, normalized size = 0.63 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (4 i \left (\tan ^{4}\left (f x +e \right )\right )+2 \left (\tan ^{5}\left (f x +e \right )\right )+6 i \left (\tan ^{2}\left (f x +e \right )\right )+\tan ^{3}\left (f x +e \right )+2 i-\tan \left (f x +e \right )\right )}{5 f \,c^{3} a \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^3/a*(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 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(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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mupad [B]  time = 5.36, size = 128, normalized size = 0.69 \[ -\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-10\,\sin \left (2\,e+2\,f\,x\right )-\sin \left (4\,e+4\,f\,x\right )+15{}\mathrm {i}\right )}{40\,a\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)^(1/2)*(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(4*e + 4*f*x)*1i - 10*si
n(2*e + 2*f*x) - sin(4*e + 4*f*x) + 15i))/(40*a*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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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