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

Optimal. Leaf size=162 \[ -\frac {32 i \sqrt {a+i a \tan (c+d x)}}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {16 i}{77 a^2 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {12 i}{77 a d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}} \]

[Out]

16/77*I/a^2/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-32/77*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/(e*sec(d*x+
c))^(1/2)+2/11*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2)+12/77*I/a/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d
*x+c))^(3/2)

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Rubi [A]  time = 0.30, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3502, 3488} \[ -\frac {32 i \sqrt {a+i a \tan (c+d x)}}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {16 i}{77 a^2 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {12 i}{77 a d (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}+\frac {2 i}{11 d (a+i a \tan (c+d x))^{5/2} \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)/11)/(d*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + ((12*I)/77)/(a*d*Sqrt[e*Sec[c + d*x]]*(a +
I*a*Tan[c + d*x])^(3/2)) + ((16*I)/77)/(a^2*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((32*I)/77)*
Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d*Sqrt[e*Sec[c + d*x]])

Rule 3488

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 3502

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}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {6 \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{11 a}\\ &=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {12 i}{77 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {24 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{77 a^2}\\ &=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {12 i}{77 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a^2 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{77 a^3}\\ &=\frac {2 i}{11 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {12 i}{77 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {16 i}{77 a^2 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{77 a^3 d \sqrt {e \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 102, normalized size = 0.63 \[ \frac {i \sec ^3(c+d x) (-22 i \sin (c+d x)+42 i \sin (3 (c+d x))-55 \cos (c+d x)+35 \cos (3 (c+d x)))}{154 a^2 d (\tan (c+d x)-i)^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/154)*Sec[c + d*x]^3*(-55*Cos[c + d*x] + 35*Cos[3*(c + d*x)] - (22*I)*Sin[c + d*x] + (42*I)*Sin[3*(c + d*x)
]))/(a^2*d*Sqrt[e*Sec[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [A]  time = 2.47, size = 89, normalized size = 0.55 \[ \frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-77 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 110 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-\frac {11}{2} i \, d x - \frac {11}{2} i \, c\right )}}{308 \, a^{3} d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/308*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-77*I*e^(8*I*d*x + 8*I*c) + 110*I*e
^(4*I*d*x + 4*I*c) + 40*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-11/2*I*d*x - 11/2*I*c)/(a^3*d*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.19, size = 140, normalized size = 0.86 \[ \frac {2 \cos \left (d x +c \right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (28 i \left (\cos ^{6}\left (d x +c \right )\right )+28 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-9 i \left (\cos ^{4}\left (d x +c \right )\right )+5 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 i \left (\cos ^{2}\left (d x +c \right )\right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-16 i\right )}{77 d e \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77/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)*(28*I*cos(d*x+c)^6+28*co
s(d*x+c)^5*sin(d*x+c)-9*I*cos(d*x+c)^4+5*cos(d*x+c)^3*sin(d*x+c)+2*I*cos(d*x+c)^2+8*cos(d*x+c)*sin(d*x+c)-16*I
)/e/a^3

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maxima [A]  time = 1.05, size = 178, normalized size = 1.10 \[ \frac {7 i \, \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 i \, \cos \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 i \, \cos \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) - 77 i \, \cos \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 7 \, \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 \, \sin \left (\frac {7}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {3}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right ) + 77 \, \sin \left (\frac {1}{11} \, \arctan \left (\sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ), \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )\right )}{308 \, a^{\frac {5}{2}} d \sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/308*(7*I*cos(11/2*d*x + 11/2*c) + 33*I*cos(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 7
7*I*cos(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 77*I*cos(1/11*arctan2(sin(11/2*d*x + 1
1/2*c), cos(11/2*d*x + 11/2*c))) + 7*sin(11/2*d*x + 11/2*c) + 33*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(
11/2*d*x + 11/2*c))) + 77*sin(3/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 77*sin(1/11*arct
an2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))))/(a^(5/2)*d*sqrt(e))

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mupad [B]  time = 4.51, size = 118, normalized size = 0.73 \[ \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (154\,\sin \left (c+d\,x\right )+33\,\sin \left (3\,c+3\,d\,x\right )+7\,\sin \left (5\,c+5\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\,33{}\mathrm {i}+\cos \left (5\,c+5\,d\,x\right )\,7{}\mathrm {i}\right )}{308\,a^2\,d\,e\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((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)*(154*sin(c + d*x) + cos(3*c + 3*d*x)*33i + cos(5*c + 5*d*x)*7i + 33*sin(3*c + 3*d*x) +
7*sin(5*c + 5*d*x)))/(308*a^2*d*e*((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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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