3.185 \(\int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=123 \[ -\frac {6 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {6 a e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d}+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {2 a e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d} \]

[Out]

2/7*I*a*(e*sec(d*x+c))^(7/2)/d+2/5*a*e*(e*sec(d*x+c))^(5/2)*sin(d*x+c)/d-6/5*a*e^4*(cos(1/2*d*x+1/2*c)^2)^(1/2
)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)+6/5*a*e^3*s
in(d*x+c)*(e*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3486, 3768, 3771, 2639} \[ \frac {6 a e^3 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d}-\frac {6 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {2 a e \sin (c+d x) (e \sec (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(-6*a*e^4*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (((2*I)/7)*a*(e*Sec[c + d
*x])^(7/2))/d + (6*a*e^3*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a*e*(e*Sec[c + d*x])^(5/2)*Sin[c + d*x]
)/(5*d)

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3486

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int (e \sec (c+d x))^{7/2} (a+i a \tan (c+d x)) \, dx &=\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+a \int (e \sec (c+d x))^{7/2} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {2 a e (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac {1}{5} \left (3 a e^2\right ) \int (e \sec (c+d x))^{3/2} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {6 a e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a e (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {1}{5} \left (3 a e^4\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {6 a e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a e (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac {\left (3 a e^4\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {6 a e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i a (e \sec (c+d x))^{7/2}}{7 d}+\frac {6 a e^3 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a e (e \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 2.21, size = 156, normalized size = 1.27 \[ \frac {a e e^{-i d x} (\cos (d x)-i \sin (d x)) (e \sec (c+d x))^{5/2} (\cos (c+3 d x)+i \sin (c+3 d x)) \left (7 i e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-28 i \cos (2 (c+d x))+27 \tan (c+d x)+7 \sin (3 (c+d x)) \sec (c+d x)-36 i\right )}{70 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(a*e*(e*Sec[c + d*x])^(5/2)*(Cos[d*x] - I*Sin[d*x])*(Cos[c + 3*d*x] + I*Sin[c + 3*d*x])*(-36*I - (28*I)*Cos[2*
(c + d*x)] + ((7*I)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^
((2*I)*(c + d*x)) + 7*Sec[c + d*x]*Sin[3*(c + d*x)] + 27*Tan[c + d*x]))/(70*d*E^(I*d*x))

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-42 i \, a e^{3} e^{\left (7 i \, d x + 7 i \, c\right )} - 154 i \, a e^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 46 i \, a e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 14 i \, a e^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (\frac {3 i \, \sqrt {2} a e^{3} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, d}, x\right )}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/35*(sqrt(2)*(-42*I*a*e^3*e^(7*I*d*x + 7*I*c) - 154*I*a*e^3*e^(5*I*d*x + 5*I*c) - 46*I*a*e^3*e^(3*I*d*x + 3*I
*c) - 14*I*a*e^3*e^(I*d*x + I*c))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 35*(d*e^(6*I*d*x
+ 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*integral(3/5*I*sqrt(2)*a*e^3*sqrt(e/(e^(2*I
*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/d, x))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2
*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a), x)

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maple [B]  time = 0.94, size = 365, normalized size = 2.97 \[ -\frac {2 a \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (21 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-21 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+21 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-21 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+21 \left (\cos ^{4}\left (d x +c \right )\right )-14 \left (\cos ^{3}\left (d x +c \right )\right )-5 i \sin \left (d x +c \right )-7 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}}}{35 d \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c)),x)

[Out]

-2/35*a/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(21*I*sin(d*x+c)*cos(d*x+c)^4*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I*sin(d*x+c)*cos(d*x+c)^4*(1/(1+cos(d*x+c
)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+21*I*sin(d*x+c)*cos(d*x+
c)^3*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-21*I
*sin(d*x+c)*cos(d*x+c)^3*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c)
)/sin(d*x+c),I)+21*cos(d*x+c)^4-14*cos(d*x+c)^3-5*I*sin(d*x+c)-7*cos(d*x+c))*(e/cos(d*x+c))^(7/2)/sin(d*x+c)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(7/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i),x)

[Out]

int((e/cos(c + d*x))^(7/2)*(a + a*tan(c + d*x)*1i), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(7/2)*(a+I*a*tan(d*x+c)),x)

[Out]

Timed out

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