3.11.14 \(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx\) [1014]

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

[Out]

1/9*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(9/2)+2/63*I*(c-I*c*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e
))^(7/2)+2/315*I*(c-I*c*tan(f*x+e))^(5/2)/a^2/f/(a+I*a*tan(f*x+e))^(5/2)

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Rubi [A]
time = 0.10, 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 = {3604, 47, 37} \begin {gather*} \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}} \end {gather*}

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 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 47

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 3604

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))^{9/2}} \, dx &=\frac {(a c) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 3.62, size = 112, normalized size = 0.82 \begin {gather*} \frac {c^2 \sec ^4(e+f x) (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)}} \end {gather*}

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 = 0.39, size = 99, normalized size = 0.73

method	result	size
derivativedivides	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i \left (\tan ^{3}\left (f x +e \right )\right )-33 i \tan \left (f x +e \right )+12 \left (\tan ^{2}\left (f x +e \right )\right )+47\right )}{315 f \,a^{5} \left (-\tan \left (f x +e \right )+i\right )^{6}}\)	\(99\)
default	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i \left (\tan ^{3}\left (f x +e \right )\right )-33 i \tan \left (f x +e \right )+12 \left (\tan ^{2}\left (f x +e \right )\right )+47\right )}{315 f \,a^{5} \left (-\tan \left (f x +e \right )+i\right )^{6}}\)	\(99\)

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,method=_RETURNVERBOSE)

[Out]

-1/315*I/f*(-c*(I*tan(f*x+e)-1))^(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)^3-3
3*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 = 0.64, size = 158, normalized size = 1.16 \begin {gather*} \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 {gather*}

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 = 1.11, size = 105, normalized size = 0.77 \begin {gather*} \frac {{\left (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 {gather*}

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*(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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3878 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 7.33, size = 184, normalized size = 1.35 \begin {gather*} \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 (63\,\sin \left (4\,e+4\,f\,x\right )+153\,\sin \left (6\,e+6\,f\,x\right )+125\,\sin \left (8\,e+8\,f\,x\right )+35\,\sin \left (10\,e+10\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,153{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,125{}\mathrm {i}+\cos \left (10\,e+10\,f\,x\right )\,35{}\mathrm {i}\right )}{2520\,a^5\,f} \end {gather*}

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)^(9/2),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)*63i + cos(6*e + 6*f*x)*153i + cos(8*e
+ 8*f*x)*125i + cos(10*e + 10*f*x)*35i + 63*sin(4*e + 4*f*x) + 153*sin(6*e + 6*f*x) + 125*sin(8*e + 8*f*x) +
35*sin(10*e + 10*f*x)))/(2520*a^5*f)

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