3.5.5 \(\int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{5/2}} \, dx\) [405]

Optimal. Leaf size=81 \[ -\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

-4/5*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/e^2/(e*sec(d*x+c))^(1/2)-2/5*I*(a+I*a*tan(d*x+c))^(3/2)/d/(e*sec(d*x+c))^(
5/2)

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Rubi [A]
time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3578, 3569} \begin {gather*} -\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/5)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*e^2*Sqrt[e*Sec[c + d*x]]) - (((2*I)/5)*(a + I*a*Tan[c + d*x])^(3/
2))/(d*(e*Sec[c + d*x])^(5/2))

Rule 3569

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[a*((m + n)/(m*d^2)), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}}+\frac {(2 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{5 d (e \sec (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 84, normalized size = 1.04 \begin {gather*} -\frac {2 a (\cos (d x)-i \sin (d x)) (\cos (c+2 d x)+i \sin (c+2 d x)) (3 i+2 \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{5 d e (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(Cos[d*x] - I*Sin[d*x])*(Cos[c + 2*d*x] + I*Sin[c + 2*d*x])*(3*I + 2*Tan[c + d*x])*Sqrt[a + I*a*Tan[c +
d*x]])/(5*d*e*(e*Sec[c + d*x])^(3/2))

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Maple [A]
time = 0.79, size = 86, normalized size = 1.06

method	result	size
risch	\(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+5\right )}{5 e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(74\)
default	\(-\frac {2 \left (i \left (\cos ^{2}\left (d x +c \right )\right )-\sin \left (d x +c \right ) \cos \left (d x +c \right )+2 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (d x +c \right )\right ) a}{5 d \,e^{5}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/d*(I*cos(d*x+c)^2-sin(d*x+c)*cos(d*x+c)+2*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c)
)^(5/2)*cos(d*x+c)^3/e^5*a

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Maxima [A]
time = 0.55, size = 58, normalized size = 0.72 \begin {gather*} \frac {{\left (-i \, a \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, a \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} e^{\left (-\frac {5}{2}\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/5*(-I*a*cos(5/2*d*x + 5/2*c) - 5*I*a*cos(1/2*d*x + 1/2*c) + a*sin(5/2*d*x + 5/2*c) + 5*a*sin(1/2*d*x + 1/2*c
))*sqrt(a)*e^(-5/2)/d

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Fricas [A]
time = 0.38, size = 73, normalized size = 0.90 \begin {gather*} \frac {{\left (-i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c - \frac {5}{2}\right )}}{5 \, d \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/5*(-I*a*e^(4*I*d*x + 4*I*c) - 6*I*a*e^(2*I*d*x + 2*I*c) - 5*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*
d*x + 1/2*I*c - 5/2)/(d*sqrt(e^(2*I*d*x + 2*I*c) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)/(e*sec(d*x+c))**(5/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)/(e*sec(c + d*x))**(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((a+I*a*tan(d*x+c))^(3/2)/(e*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.71, size = 102, normalized size = 1.26 \begin {gather*} -\frac {a\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,11{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{10\,d\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/(e/cos(c + d*x))^(5/2),x)

[Out]

-(a*(e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(co
s(c + d*x)*11i - sin(c + d*x) + cos(3*c + 3*d*x)*1i - sin(3*c + 3*d*x)))/(10*d*e^3)

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