3.5.21 \(\int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [421]

Optimal. Leaf size=165 \[ \frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}} \]

[Out]

2/7*I/d/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2)+16/35*I/d/e^2/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1
/2)-12/35*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*sec(d*x+c))^(5/2)-32/35*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/e^2/(e*sec(
d*x+c))^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3583, 3578, 3569} \begin {gather*} -\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/7)/(d*(e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((16*I)/35)/(d*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt
[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(5/2)) - (((32*I)/35)
*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*e^2*Sqrt[e*Sec[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 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]

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 {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {6 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{7 a}\\ &=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {24 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{35 e^2}\\ &=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{35 a e^2}\\ &=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 79, normalized size = 0.48 \begin {gather*} -\frac {i (17+\cos (2 (c+d x))+3 i \sec (c+d x) \sin (3 (c+d x))+35 i \tan (c+d x))}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-1/35*I)*(17 + Cos[2*(c + d*x)] + (3*I)*Sec[c + d*x]*Sin[3*(c + d*x)] + (35*I)*Tan[c + d*x]))/(d*e^2*Sqrt[e*
Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.89, size = 115, normalized size = 0.70

method	result	size
default	\(\frac {2 \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (5 i \left (\cos ^{4}\left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+2 i \left (\cos ^{2}\left (d x +c \right )\right )+8 \sin \left (d x +c \right ) \cos \left (d x +c \right )-16 i\right )}{35 d \,e^{5} a}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35/d*(e/cos(d*x+c))^(5/2)*cos(d*x+c)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(5*I*cos(d*x+c)^4+5*si
n(d*x+c)*cos(d*x+c)^3+2*I*cos(d*x+c)^2+8*sin(d*x+c)*cos(d*x+c)-16*I)/e^5/a

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Maxima [A]
time = 0.56, size = 177, normalized size = 1.07 \begin {gather*} \frac {{\left (5 i \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 7 i \, \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 i \, \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 105 i \, \cos \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} e^{\left (-\frac {5}{2}\right )}}{140 \, \sqrt {a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*(5*I*cos(7/2*d*x + 7/2*c) - 7*I*cos(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35*I*cos(
3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 105*I*cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2
*d*x + 7/2*c))) + 5*sin(7/2*d*x + 7/2*c) + 7*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 35
*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(
7/2*d*x + 7/2*c))))*e^(-5/2)/(sqrt(a)*d)

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Fricas [A]
time = 0.39, size = 94, normalized size = 0.57 \begin {gather*} \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 112 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c - \frac {5}{2}\right )}}{140 \, a 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(1/(e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/140*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-7*I*e^(8*I*d*x + 8*I*c) - 112*I*e^(6*I*d*x + 6*I*c) - 70*I*e^(4*I*d*
x + 4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7/2*I*d*x - 7/2*I*c - 5/2)/(a*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 {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 4.21, size = 101, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (-\sin \left (c+d\,x\right )-\frac {3\,\sin \left (3\,c+3\,d\,x\right )}{35}+\frac {\cos \left (c+d\,x\right )\,1{}\mathrm {i}}{2}+\frac {\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{70}\right )}{d\,e^3\,\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(1/((e/cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

-((e/cos(c + d*x))^(1/2)*((cos(c + d*x)*1i)/2 - sin(c + d*x) + (cos(3*c + 3*d*x)*1i)/70 - (3*sin(3*c + 3*d*x))
/35))/(d*e^3*((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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