3.5.34 \(\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [434]

Optimal. Leaf size=121 \[ \frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

16/45*I*(e*sec(d*x+c))^(1/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+2/9*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(5
/2)+8/45*I*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.16, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3583, 3569} \begin {gather*} \frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/9)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((8*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + (((16*I)/45)*Sqrt[e*Sec[c + d*x]])/(a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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 3583

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

Rubi steps

\begin {align*} \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{9 a}\\ &=\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {8 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{45 a^2}\\ &=\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 85, normalized size = 0.70 \begin {gather*} -\frac {i \sec ^2(c+d x) \sqrt {e \sec (c+d x)} (9+25 \cos (2 (c+d x))+20 i \sin (2 (c+d x)))}{45 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/45*I)*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(9 + 25*Cos[2*(c + d*x)] + (20*I)*Sin[2*(c + d*x)]))/(a^2*d*(-I
+ Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.90, size = 128, normalized size = 1.06

method	result	size
default	\(\frac {2 i \cos \left (d x +c \right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-20 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+20 \left (\cos ^{5}\left (d x +c \right )\right )-3 i \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-7 \left (\cos ^{3}\left (d x +c \right )\right )-8 i \sin \left (d x +c \right )+4 \cos \left (d x +c \right )\right )}{45 d \,a^{3}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/45*I/d*cos(d*x+c)*(e/cos(d*x+c))^(1/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-20*I*cos(d*x+c)^4*si
n(d*x+c)+20*cos(d*x+c)^5-3*I*cos(d*x+c)^2*sin(d*x+c)-7*cos(d*x+c)^3-8*I*sin(d*x+c)+4*cos(d*x+c))/a^3

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Maxima [A]
time = 0.54, size = 129, normalized size = 1.07 \begin {gather*} \frac {{\left (5 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )\right )} e^{\frac {1}{2}}}{90 \, a^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/90*(5*I*cos(9/2*d*x + 9/2*c) + 18*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*I*cos(
1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 5*sin(9/2*d*x + 9/2*c) + 18*sin(5/9*arctan2(sin(9/2
*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))*e^(1/
2)/(a^(5/2)*d)

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Fricas [A]
time = 0.38, size = 88, normalized size = 0.73 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\frac {1}{2}} + 45 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {1}{2}\right )} + 63 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {1}{2}\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{90 \, a^{3} 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((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/90*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(5*I*e^(1/2) + 45*I*e^(6*I*d*x + 6*I*c + 1/2) + 63*I*e^(4*I*d*x + 4*I*c
+ 1/2) + 23*I*e^(2*I*d*x + 2*I*c + 1/2))*e^(-9/2*I*d*x - 9/2*I*c)/(a^3*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 {\sqrt {e \sec {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sqrt(e*sec(c + d*x))/(I*a*(tan(c + d*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((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(e^(1/2)*sqrt(sec(d*x + c))/(I*a*tan(d*x + c) + a)^(5/2), x)

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Mupad [B]
time = 4.22, size = 109, normalized size = 0.90 \begin {gather*} \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,18{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}+18\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (4\,c+4\,d\,x\right )+45{}\mathrm {i}\right )}{90\,a^2\,d\,\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*18i + cos(4*c + 4*d*x)*5i + 18*sin(2*c + 2*d*x) + 5*sin(4*c + 4*d*x)
+ 45i))/(90*a^2*d*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))

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