3.1013 \(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {i (c-i c \tan (e+f x))^{5/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 f (a+i a \tan (e+f x))^{7/2}} \]

[Out]

1/7*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(7/2)+1/35*I*(c-I*c*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e
))^(5/2)

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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} \[ \frac {i (c-i c \tan (e+f x))^{5/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 f (a+i a \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(7/2),x]

[Out]

((I/7)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(7/2)) + ((I/35)*(c - I*c*Tan[e + f*x])^(5/2))/
(a*f*(a + I*a*Tan[e + f*x])^(5/2))

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 {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i (c-i c \tan (e+f x))^{5/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {c \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac {i (c-i c \tan (e+f x))^{5/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{35 a f (a+i a \tan (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 6.52, size = 100, normalized size = 1.11 \[ -\frac {i c^2 (\tan (e+f x)-6 i) \sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{35 a^3 f (\tan (e+f x)-i)^3 \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(7/2),x]

[Out]

((-1/35*I)*c^2*Sec[e + f*x]^2*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*(-6*I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e
+ f*x]])/(a^3*f*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [A]  time = 0.59, size = 85, normalized size = 0.94 \[ \frac {{\left (7 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-7 i \, f x - 7 i \, e\right )}}{70 \, a^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/70*(7*I*c^2*e^(4*I*f*x + 4*I*e) + 12*I*c^2*e^(2*I*f*x + 2*I*e) + 5*I*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*
sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-7*I*f*x - 7*I*e)/(a^4*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(7/2), x)

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maple [A]  time = 0.27, size = 87, normalized size = 0.97 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (5 i \tan \left (f x +e \right )-\left (\tan ^{2}\left (f x +e \right )\right )-6\right )}{35 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x)

[Out]

1/35/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^4*(1+tan(f*x+e)^2)*(5*I*tan(f*x+e)-tan(f*
x+e)^2-6)/(-tan(f*x+e)+I)^5

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maxima [A]  time = 0.63, size = 94, normalized size = 1.04 \[ \frac {{\left (5 i \, c^{2} \cos \left (7 \, f x + 7 \, e\right ) + 7 i \, c^{2} \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right ) + 5 \, c^{2} \sin \left (7 \, f x + 7 \, e\right ) + 7 \, c^{2} \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right )\right )} \sqrt {c}}{70 \, a^{\frac {7}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

1/70*(5*I*c^2*cos(7*f*x + 7*e) + 7*I*c^2*cos(5/7*arctan2(sin(7*f*x + 7*e), cos(7*f*x + 7*e))) + 5*c^2*sin(7*f*
x + 7*e) + 7*c^2*sin(5/7*arctan2(sin(7*f*x + 7*e), cos(7*f*x + 7*e))))*sqrt(c)/(a^(7/2)*f)

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mupad [B]  time = 6.47, size = 161, normalized size = 1.79 \[ \frac {c^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (7\,\sin \left (4\,e+4\,f\,x\right )+12\,\sin \left (6\,e+6\,f\,x\right )+5\,\sin \left (8\,e+8\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )\,7{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,12{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,5{}\mathrm {i}\right )}{140\,a^4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^(5/2)/(a + a*tan(e + f*x)*1i)^(7/2),x)

[Out]

(c^2*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*((c*(cos(2*e + 2*f*x) - s
in(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(4*e + 4*f*x)*7i + cos(6*e + 6*f*x)*12i + cos(8*e +
8*f*x)*5i + 7*sin(4*e + 4*f*x) + 12*sin(6*e + 6*f*x) + 5*sin(8*e + 8*f*x)))/(140*a^4*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(7/2),x)

[Out]

Timed out

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