3.11.15 \(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{11/2}} \, dx\) [1015]

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

[Out]

1/11*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(11/2)+1/33*I*(c-I*c*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x
+e))^(9/2)+2/231*I*(c-I*c*tan(f*x+e))^(5/2)/a^2/f/(a+I*a*tan(f*x+e))^(7/2)+2/1155*I*(c-I*c*tan(f*x+e))^(5/2)/a
^3/f/(a+I*a*tan(f*x+e))^(5/2)

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Rubi [A]
time = 0.11, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 = {3604, 47, 37} \begin {gather*} \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}} \end {gather*}

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 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))^{11/2}} \, dx &=\frac {(a c) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 5.32, size = 128, normalized size = 0.70 \begin {gather*} \frac {c^2 \sec ^4(e+f x) (\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)}} \end {gather*}

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 = 0.37, size = 110, normalized size = 0.60

method	result	size
derivativedivides	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i \left (\tan ^{4}\left (f x +e \right )\right )-45 i \left (\tan ^{2}\left (f x +e \right )\right )+14 \left (\tan ^{3}\left (f x +e \right )\right )-152 i-91 \tan \left (f x +e \right )\right )}{1155 f \,a^{6} \left (-\tan \left (f x +e \right )+i\right )^{7}}\)	\(110\)
default	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i \left (\tan ^{4}\left (f x +e \right )\right )-45 i \left (\tan ^{2}\left (f x +e \right )\right )+14 \left (\tan ^{3}\left (f x +e \right )\right )-152 i-91 \tan \left (f x +e \right )\right )}{1155 f \,a^{6} \left (-\tan \left (f x +e \right )+i\right )^{7}}\)	\(110\)

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,method=_RETURNVERBOSE)

[Out]

1/1155*I/f*(-c*(I*tan(f*x+e)-1))^(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)^4-4
5*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 = 0.57, size = 216, normalized size = 1.19 \begin {gather*} \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 {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))^(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 = 0.99, size = 120, normalized size = 0.66 \begin {gather*} \frac {{\left (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 {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))^(11/2),x, algorithm="fricas")

[Out]

1/9240*(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)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-1
1*I*f*x - 11*I*e)/(a^6*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(11/2),x)

[Out]

Timed out

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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))^(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)

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Mupad [B]
time = 8.15, size = 207, normalized size = 1.14 \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 (231\,\sin \left (4\,e+4\,f\,x\right )+726\,\sin \left (6\,e+6\,f\,x\right )+880\,\sin \left (8\,e+8\,f\,x\right )+490\,\sin \left (10\,e+10\,f\,x\right )+105\,\sin \left (12\,e+12\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )\,231{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,726{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,880{}\mathrm {i}+\cos \left (10\,e+10\,f\,x\right )\,490{}\mathrm {i}+\cos \left (12\,e+12\,f\,x\right )\,105{}\mathrm {i}\right )}{18480\,a^6\,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)^(11/2),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)*231i + cos(6*e + 6*f*x)*726i + cos(8*
e + 8*f*x)*880i + cos(10*e + 10*f*x)*490i + cos(12*e + 12*f*x)*105i + 231*sin(4*e + 4*f*x) + 726*sin(6*e + 6*f
*x) + 880*sin(8*e + 8*f*x) + 490*sin(10*e + 10*f*x) + 105*sin(12*e + 12*f*x)))/(18480*a^6*f)

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