3.1014 \(\int \frac{(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=136 \[ \frac{2 i (c-i c \tan (e+f x))^{5/2}}{315 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{63 a f (a+i a \tan (e+f x))^{7/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}} \]

[Out]

((I/9)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(9/2)) + (((2*I)/63)*(c - I*c*Tan[e + f*x])^(5/
2))/(a*f*(a + I*a*Tan[e + f*x])^(7/2)) + (((2*I)/315)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*f*(a + I*a*Tan[e + f*
x])^(5/2))

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Rubi [A]  time = 0.142366, antiderivative size = 136, normalized size of antiderivative = 1., 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} \[ \frac{2 i (c-i c \tan (e+f x))^{5/2}}{315 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{63 a f (a+i a \tan (e+f x))^{7/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(9/2),x]

[Out]

((I/9)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(9/2)) + (((2*I)/63)*(c - I*c*Tan[e + f*x])^(5/
2))/(a*f*(a + I*a*Tan[e + f*x])^(7/2)) + (((2*I)/315)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*f*(a + I*a*Tan[e + f*
x])^(5/2))

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]

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 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]

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac{(2 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 )}{9 f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{63 a f (a+i a \tan (e+f x))^{7/2}}+\frac{(2 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 )}{63 a f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{63 a f (a+i a \tan (e+f x))^{7/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{315 a^2 f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 7.52975, size = 112, normalized size = 0.82 \[ \frac{c^2 \sec ^4(e+f x) \sqrt{c-i c \tan (e+f x)} (14 i \sin (2 (e+f x))+49 \cos (2 (e+f x))+45) (\sin (2 (e+f x))+i \cos (2 (e+f x)))}{630 a^4 f (\tan (e+f x)-i)^4 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(9/2),x]

[Out]

(c^2*Sec[e + f*x]^4*(45 + 49*Cos[2*(e + f*x)] + (14*I)*Sin[2*(e + f*x)])*(I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)
])*Sqrt[c - I*c*Tan[e + f*x]])/(630*a^4*f*(-I + Tan[e + f*x])^4*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.038, size = 99, normalized size = 0.7 \begin{align*}{\frac{-{\frac{i}{315}}{c}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 2\,i \left ( \tan \left ( fx+e \right ) \right ) ^{3}-33\,i\tan \left ( fx+e \right ) +12\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+47 \right ) }{f{a}^{5} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{6}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

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))^(9/2),x)

[Out]

-1/315*I/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^5*(1+tan(f*x+e)^2)*(2*I*tan(f*x+e)^3-
33*I*tan(f*x+e)+12*tan(f*x+e)^2+47)/(-tan(f*x+e)+I)^6

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Maxima [A]  time = 2.04229, size = 200, normalized size = 1.47 \begin{align*} \frac{{\left (35 i \, c^{2} \cos \left (9 \, f x + 9 \, e\right ) + 90 i \, c^{2} \cos \left (\frac{7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 63 i \, c^{2} \cos \left (\frac{5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 35 \, c^{2} \sin \left (9 \, f x + 9 \, e\right ) + 90 \, c^{2} \sin \left (\frac{7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 63 \, c^{2} \sin \left (\frac{5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right )\right )} \sqrt{c}}{1260 \, a^{\frac{9}{2}} f} \end{align*}

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))^(9/2),x, algorithm="maxima")

[Out]

1/1260*(35*I*c^2*cos(9*f*x + 9*e) + 90*I*c^2*cos(7/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 63*I*c^2*c
os(5/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 35*c^2*sin(9*f*x + 9*e) + 90*c^2*sin(7/9*arctan2(sin(9*f
*x + 9*e), cos(9*f*x + 9*e))) + 63*c^2*sin(5/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))))*sqrt(c)/(a^(9/2)*
f)

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Fricas [A]  time = 1.64723, size = 375, normalized size = 2.76 \begin{align*} \frac{{\left (-188 i \, c^{2} e^{\left (11 i \, f x + 11 i \, e\right )} - 188 i \, c^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + 63 i \, c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 153 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 125 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 35 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 (-9 i \, f x - 9 i \, e\right )}}{1260 \, a^{5} f} \end{align*}

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))^(9/2),x, algorithm="fricas")

[Out]

1/1260*(-188*I*c^2*e^(11*I*f*x + 11*I*e) - 188*I*c^2*e^(9*I*f*x + 9*I*e) + 63*I*c^2*e^(6*I*f*x + 6*I*e) + 153*
I*c^2*e^(4*I*f*x + 4*I*e) + 125*I*c^2*e^(2*I*f*x + 2*I*e) + 35*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^(-9*I*f*x - 9*I*e)/(a^5*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{9}{2}}}\,{d x} \end{align*}

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))^(9/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(9/2), x)