3.5.12 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{5/2}} \, dx\) [412]

Optimal. Leaf size=38 \[ -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {3569} \begin {gather*} -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(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]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 38, normalized size = 1.00 \begin {gather*} -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 d (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (30 ) = 60\).
time = 0.79, size = 88, normalized size = 2.32

method	result	size
risch	\(-\frac {2 i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{5 e^{2} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(74\)
default	\(-\frac {2 \left (2 i \left (\cos ^{2}\left (d x +c \right )\right )-2 \sin \left (d x +c \right ) \cos \left (d x +c \right )-i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (d x +c \right )\right ) a^{2}}{5 d \,e^{5}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d*(2*I*cos(d*x+c)^2-2*sin(d*x+c)*cos(d*x+c)-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^2

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (28) = 56\).
time = 0.50, size = 75, normalized size = 1.97 \begin {gather*} -\frac {2 i \, a^{\frac {5}{2}} {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {5}{2}} e^{\left (-\frac {5}{2}\right )}}{5 \, d {\left (-\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (28) = 56\).
time = 0.37, size = 74, normalized size = 1.95 \begin {gather*} \frac {2 \, {\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )}\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))^(5/2)/(e*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 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((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.55, size = 104, normalized size = 2.74 \begin {gather*} -\frac {a^2\,\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 )\,1{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{5\,d\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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