3.1037 \(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=90 \[ -\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 4.51, size = 106, normalized size = 1.18 \[ \frac {a \cos (e+f x) (\cos (f x)-i \sin (f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (4 \cos (e+f x)-i \sin (e+f x)) (\sin (4 e+5 f x)-i \cos (4 e+5 f x))}{15 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*(4*Cos[e + f*x] - I*Sin[e + f*x])*((-I)*Cos[4*e + 5*f*x] + Sin[4*e + 5
*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f)

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fricas [A]  time = 0.43, size = 79, normalized size = 0.88 \[ \frac {{\left (-3 i \, a e^{\left (7 i \, f x + 7 i \, e\right )} - 8 i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - 5 i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(-3*I*a*e^(7*I*f*x + 7*I*e) - 8*I*a*e^(5*I*f*x + 5*I*e) - 5*I*a*e^(3*I*f*x + 3*I*e))*sqrt(a/(e^(2*I*f*x +
2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 71, normalized size = 0.79 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.54, size = 112, normalized size = 1.24 \[ \frac {{\left (-3 i \, a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 5 i \, a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(-3*I*a*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 5*I*a*cos(3/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e))) + 3*a*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 5*a*sin(3/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(5/2)*f)

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mupad [B]  time = 5.47, size = 136, normalized size = 1.51 \[ -\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (-5\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*((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)*5i + cos
(4*e + 4*f*x)*3i - 5*sin(2*e + 2*f*x) - 3*sin(4*e + 4*f*x)))/(30*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 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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