3.1006 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=182 \[ \frac {2 i (c-i c \tan (e+f x))^{3/2}}{315 a^3 f (a+i a \tan (e+f x))^{3/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{21 a f (a+i a \tan (e+f x))^{7/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}} \]

[Out]

1/9*I*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(9/2)+1/21*I*(c-I*c*tan(f*x+e))^(3/2)/a/f/(a+I*a*tan(f*x+e
))^(7/2)+2/105*I*(c-I*c*tan(f*x+e))^(3/2)/a^2/f/(a+I*a*tan(f*x+e))^(5/2)+2/315*I*(c-I*c*tan(f*x+e))^(3/2)/a^3/
f/(a+I*a*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.17, 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 = {3523, 45, 37} \[ \frac {2 i (c-i c \tan (e+f x))^{3/2}}{315 a^3 f (a+i a \tan (e+f x))^{3/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{21 a f (a+i a \tan (e+f x))^{7/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(9/2),x]

[Out]

((I/9)*(c - I*c*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(9/2)) + ((I/21)*(c - I*c*Tan[e + f*x])^(3/2))/
(a*f*(a + I*a*Tan[e + f*x])^(7/2)) + (((2*I)/105)*(c - I*c*Tan[e + f*x])^(3/2))/(a^2*f*(a + I*a*Tan[e + f*x])^
(5/2)) + (((2*I)/315)*(c - I*c*Tan[e + f*x])^(3/2))/(a^3*f*(a + I*a*Tan[e + f*x])^(3/2))

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))^{3/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac {c \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{21 a f (a+i a \tan (e+f x))^{7/2}}+\frac {(2 c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{21 a f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{21 a f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{105 a^2 f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{21 a f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{105 a^2 f (a+i a \tan (e+f x))^{5/2}}+\frac {2 i (c-i c \tan (e+f x))^{3/2}}{315 a^3 f (a+i a \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 5.55, size = 115, normalized size = 0.63 \[ \frac {c (\tan (e+f x)+i) \sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (140 \cos (2 (e+f x))+27 i \tan (e+f x)+35 i \sin (3 (e+f x)) \sec (e+f x)+92)}{1260 a^4 f (\tan (e+f x)-i)^4 \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(9/2),x]

[Out]

(c*Sec[e + f*x]^2*(92 + 140*Cos[2*(e + f*x)] + (35*I)*Sec[e + f*x]*Sin[3*(e + f*x)] + (27*I)*Tan[e + f*x])*(I
+ Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(1260*a^4*f*(-I + Tan[e + f*x])^4*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [A]  time = 0.51, size = 103, normalized size = 0.57 \[ \frac {{\left (105 i \, c e^{\left (8 i \, f x + 8 i \, e\right )} + 294 i \, c e^{\left (6 i \, f x + 6 i \, e\right )} + 324 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 170 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 35 i \, c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-9 i \, f x - 9 i \, e\right )}}{2520 \, a^{5} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/2520*(105*I*c*e^(8*I*f*x + 8*I*e) + 294*I*c*e^(6*I*f*x + 6*I*e) + 324*I*c*e^(4*I*f*x + 4*I*e) + 170*I*c*e^(2
*I*f*x + 2*I*e) + 35*I*c)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^(9/2), x)

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maple [A]  time = 0.36, size = 97, normalized size = 0.53 \[ \frac {i \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (2 i \left (\tan ^{3}\left (f x +e \right )\right )-33 i \tan \left (f x +e \right )+12 \left (\tan ^{2}\left (f x +e \right )\right )-58\right )}{315 f \,a^{5} \left (-\tan \left (f x +e \right )+i\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(3/2)/(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/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)^2-58)/(-tan(f*x+e)+I)^6

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maxima [A]  time = 0.69, size = 186, normalized size = 1.02 \[ \frac {{\left (35 i \, c \cos \left (9 \, f x + 9 \, e\right ) + 135 i \, c \cos \left (\frac {7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 189 i \, c \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 105 i \, c \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 35 \, c \sin \left (9 \, f x + 9 \, e\right ) + 135 \, c \sin \left (\frac {7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 189 \, c \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 105 \, c \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right )\right )} \sqrt {c}}{2520 \, a^{\frac {9}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

1/2520*(35*I*c*cos(9*f*x + 9*e) + 135*I*c*cos(7/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 189*I*c*cos(5
/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 105*I*c*cos(1/3*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e)))
+ 35*c*sin(9*f*x + 9*e) + 135*c*sin(7/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 189*c*sin(5/9*arctan2(
sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 105*c*sin(1/3*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))))*sqrt(c)/(a^
(9/2)*f)

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mupad [B]  time = 7.30, size = 205, normalized size = 1.13 \[ \frac {c\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (105\,\sin \left (2\,e+2\,f\,x\right )+294\,\sin \left (4\,e+4\,f\,x\right )+324\,\sin \left (6\,e+6\,f\,x\right )+170\,\sin \left (8\,e+8\,f\,x\right )+35\,\sin \left (10\,e+10\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,105{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,294{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,324{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,170{}\mathrm {i}+\cos \left (10\,e+10\,f\,x\right )\,35{}\mathrm {i}\right )}{5040\,a^5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*tan(e + f*x)*1i)^(3/2)/(a + a*tan(e + f*x)*1i)^(9/2),x)

[Out]

(c*((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) - sin
(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(cos(2*e + 2*f*x)*105i + cos(4*e + 4*f*x)*294i + cos(6*e
+ 6*f*x)*324i + cos(8*e + 8*f*x)*170i + cos(10*e + 10*f*x)*35i + 105*sin(2*e + 2*f*x) + 294*sin(4*e + 4*f*x) +
324*sin(6*e + 6*f*x) + 170*sin(8*e + 8*f*x) + 35*sin(10*e + 10*f*x)))/(5040*a^5*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))**(3/2)/(a+I*a*tan(f*x+e))**(9/2),x)

[Out]

Timed out

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