3.1015 \(\int \frac{(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=182 \[ \frac{2 i (c-i c \tan (e+f x))^{5/2}}{1155 a^3 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{231 a^2 f (a+i a \tan (e+f x))^{7/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{33 a f (a+i a \tan (e+f x))^{9/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}} \]

[Out]

((I/11)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(11/2)) + ((I/33)*(c - I*c*Tan[e + f*x])^(5/2)
)/(a*f*(a + I*a*Tan[e + f*x])^(9/2)) + (((2*I)/231)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*f*(a + I*a*Tan[e + f*x]
)^(7/2)) + (((2*I)/1155)*(c - I*c*Tan[e + f*x])^(5/2))/(a^3*f*(a + I*a*Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.164411, antiderivative size = 182, normalized size of antiderivative = 1., 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} \[ \frac{2 i (c-i c \tan (e+f x))^{5/2}}{1155 a^3 f (a+i a \tan (e+f x))^{5/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{231 a^2 f (a+i a \tan (e+f x))^{7/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{33 a f (a+i a \tan (e+f x))^{9/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(11/2),x]

[Out]

((I/11)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(11/2)) + ((I/33)*(c - I*c*Tan[e + f*x])^(5/2)
)/(a*f*(a + I*a*Tan[e + f*x])^(9/2)) + (((2*I)/231)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*f*(a + I*a*Tan[e + f*x]
)^(7/2)) + (((2*I)/1155)*(c - I*c*Tan[e + f*x])^(5/2))/(a^3*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))^{11/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}}+\frac{(3 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 )}{11 f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{33 a 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 )}{33 a f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{33 a f (a+i a \tan (e+f x))^{9/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{231 a^2 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 )}{231 a^2 f}\\ &=\frac{i (c-i c \tan (e+f x))^{5/2}}{11 f (a+i a \tan (e+f x))^{11/2}}+\frac{i (c-i c \tan (e+f x))^{5/2}}{33 a f (a+i a \tan (e+f x))^{9/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{231 a^2 f (a+i a \tan (e+f x))^{7/2}}+\frac{2 i (c-i c \tan (e+f x))^{5/2}}{1155 a^3 f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 11.1354, size = 128, normalized size = 0.7 \[ \frac{c^2 \sec ^4(e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x))) (336 \cos (2 (e+f x))+55 i \tan (e+f x)+63 i \sin (3 (e+f x)) \sec (e+f x)+272)}{4620 a^5 f (\tan (e+f x)-i)^5 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(11/2),x]

[Out]

(c^2*Sec[e + f*x]^4*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*(272 + 336*Cos[2*(e + f*x)] + (63*I)*Sec[e + f*x]*
Sin[3*(e + f*x)] + (55*I)*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(4620*a^5*f*(-I + Tan[e + f*x])^5*Sqrt[a +
I*a*Tan[e + f*x]])

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Maple [A]  time = 0.043, size = 110, normalized size = 0.6 \begin{align*}{\frac{{\frac{i}{1155}}{c}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 2\,i \left ( \tan \left ( fx+e \right ) \right ) ^{4}-45\,i \left ( \tan \left ( fx+e \right ) \right ) ^{2}+14\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}-152\,i-91\,\tan \left ( fx+e \right ) \right ) }{f{a}^{6} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{7}}\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))^(11/2),x)

[Out]

1/1155*I/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^6*(1+tan(f*x+e)^2)*(2*I*tan(f*x+e)^4-
45*I*tan(f*x+e)^2+14*tan(f*x+e)^3-152*I-91*tan(f*x+e))/(-tan(f*x+e)+I)^7

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Maxima [A]  time = 2.03793, size = 273, normalized size = 1.5 \begin{align*} \frac{{\left (105 i \, c^{2} \cos \left (11 \, f x + 11 \, e\right ) + 385 i \, c^{2} \cos \left (\frac{9}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right ) + 495 i \, c^{2} \cos \left (\frac{7}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right ) + 231 i \, c^{2} \cos \left (\frac{5}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right ) + 105 \, c^{2} \sin \left (11 \, f x + 11 \, e\right ) + 385 \, c^{2} \sin \left (\frac{9}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right ) + 495 \, c^{2} \sin \left (\frac{7}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right ) + 231 \, c^{2} \sin \left (\frac{5}{11} \, \arctan \left (\sin \left (11 \, f x + 11 \, e\right ), \cos \left (11 \, f x + 11 \, e\right )\right )\right )\right )} \sqrt{c}}{9240 \, a^{\frac{11}{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))^(11/2),x, algorithm="maxima")

[Out]

1/9240*(105*I*c^2*cos(11*f*x + 11*e) + 385*I*c^2*cos(9/11*arctan2(sin(11*f*x + 11*e), cos(11*f*x + 11*e))) + 4
95*I*c^2*cos(7/11*arctan2(sin(11*f*x + 11*e), cos(11*f*x + 11*e))) + 231*I*c^2*cos(5/11*arctan2(sin(11*f*x + 1
1*e), cos(11*f*x + 11*e))) + 105*c^2*sin(11*f*x + 11*e) + 385*c^2*sin(9/11*arctan2(sin(11*f*x + 11*e), cos(11*
f*x + 11*e))) + 495*c^2*sin(7/11*arctan2(sin(11*f*x + 11*e), cos(11*f*x + 11*e))) + 231*c^2*sin(5/11*arctan2(s
in(11*f*x + 11*e), cos(11*f*x + 11*e))))*sqrt(c)/(a^(11/2)*f)

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Fricas [A]  time = 1.47361, size = 429, normalized size = 2.36 \begin{align*} \frac{{\left (-1216 i \, c^{2} e^{\left (13 i \, f x + 13 i \, e\right )} - 1216 i \, c^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + 231 i \, c^{2} e^{\left (8 i \, f x + 8 i \, e\right )} + 726 i \, c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 880 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 490 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 105 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 (-11 i \, f x - 11 i \, e\right )}}{9240 \, a^{6} 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))^(11/2),x, algorithm="fricas")

[Out]

1/9240*(-1216*I*c^2*e^(13*I*f*x + 13*I*e) - 1216*I*c^2*e^(11*I*f*x + 11*I*e) + 231*I*c^2*e^(8*I*f*x + 8*I*e) +
726*I*c^2*e^(6*I*f*x + 6*I*e) + 880*I*c^2*e^(4*I*f*x + 4*I*e) + 490*I*c^2*e^(2*I*f*x + 2*I*e) + 105*I*c^2)*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-11*I*f*x - 11*I*e)/(a^6*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))**(11/2),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{11}{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))^(11/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(11/2), x)