3.1038 \(\int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 136, 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 = {3523, 45, 37} \[ -\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 3.64, size = 102, normalized size = 0.75 \[ \frac {\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} \left (19 \cos (e+f x)+9 \cos (3 (e+f x))-24 i \sin (e+f x) \cos ^2(e+f x)\right ) (\sin (3 (e+f x))-i \cos (3 (e+f x)))}{60 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*Cos[e + f*x] + 9*Cos[3*(e + f*x)] - (24*I)*Cos[e + f*x]^2*Sin[e + f*x])*((-I)*Cos[3*(e + f*x)] + Sin[3*(e
+ f*x)])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(60*c^3*f)

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fricas [A]  time = 0.48, size = 87, normalized size = 0.64 \[ \frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (7 i \, f x + 7 i \, e\right )} - 13 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 25 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 15 i \, e^{\left (i \, f x + i \, e\right )}\right )}}{60 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*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^(7*I*f*x + 7*I*e) - 13*I*e^(5
*I*f*x + 5*I*e) - 25*I*e^(3*I*f*x + 3*I*e) - 15*I*e^(I*f*x + I*e))/(c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 = 83, normalized size = 0.61 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (8 i \left (\tan ^{2}\left (f x +e \right )\right )+2 \left (\tan ^{3}\left (f x +e \right )\right )-7 i-13 \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^3*(8*I*tan(f*x+e)^2+2*tan(f*x+e)^3-7*I-13*tan
(f*x+e))/(tan(f*x+e)+I)^4

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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((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 = 5.48, size = 137, normalized size = 1.01 \[ -\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-10\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+15{}\mathrm {i}\right )}{60\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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(2*e + 2*f*x)*10i + cos(
4*e + 4*f*x)*3i - 10*sin(2*e + 2*f*x) - 3*sin(4*e + 4*f*x) + 15i))/(60*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 \[ \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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