3.11.4 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1004]

Optimal. Leaf size=90 \[ \frac {i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

1/5*I*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(5/2)+1/15*I*(c-I*c*tan(f*x+e))^(3/2)/a/f/(a+I*a*tan(f*x+e
))^(3/2)

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Rubi [A]
time = 0.09, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {i (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(5/2),x]

[Out]

((I/5)*(c - I*c*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(5/2)) + ((I/15)*(c - I*c*Tan[e + f*x])^(3/2))/
(a*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 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))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {c \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.65, size = 79, normalized size = 0.88 \begin {gather*} \frac {c (1-i \tan (e+f x)) (-4 i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - I*c*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(5/2),x]

[Out]

(c*(1 - I*Tan[e + f*x])*(-4*I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*a^2*f*(-I + Tan[e + f*x])^2*Sqrt
[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.35, size = 75, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i-\tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i-\tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*c*(1+tan(f*x+e)^2)*(4*I-tan(f*x+e))/(-tan(f*
x+e)+I)^4

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Maxima [A]
time = 0.61, size = 92, normalized size = 1.02 \begin {gather*} \frac {{\left (3 i \, c \cos \left (5 \, f x + 5 \, e\right ) + 5 i \, c \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 3 \, c \sin \left (5 \, f x + 5 \, e\right ) + 5 \, c \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )\right )} \sqrt {c}}{30 \, a^{\frac {5}{2}} f} \end {gather*}

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))^(5/2),x, algorithm="maxima")

[Out]

1/30*(3*I*c*cos(5*f*x + 5*e) + 5*I*c*cos(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))) + 3*c*sin(5*f*x + 5*
e) + 5*c*sin(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))))*sqrt(c)/(a^(5/2)*f)

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Fricas [A]
time = 1.13, size = 84, normalized size = 0.93 \begin {gather*} \frac {{\left (5 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 3 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 (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \end {gather*}

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))^(5/2),x, algorithm="fricas")

[Out]

1/30*(5*I*c*e^(4*I*f*x + 4*I*e) + 8*I*c*e^(2*I*f*x + 2*I*e) + 3*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^(-5*I*f*x - 5*I*e)/(a^3*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

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))**(5/2),x)

[Out]

Integral((-I*c*(tan(e + f*x) + I))**(3/2)/(I*a*(tan(e + f*x) - I))**(5/2), x)

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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))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^(5/2), x)

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Mupad [B]
time = 6.10, size = 159, normalized size = 1.77 \begin {gather*} \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 (5\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,8{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,a^3\,f} \end {gather*}

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)^(5/2),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)*5i + cos(4*e + 4*f*x)*8i + cos(6*e + 6*
f*x)*3i + 5*sin(2*e + 2*f*x) + 8*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x)))/(60*a^3*f)

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