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

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

[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)

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Rubi [A]  time = 0.0905456, antiderivative size = 123, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 2.13883, size = 156, normalized size = 1.27 \[ \frac{a e e^{-i d x} (\cos (d x)-i \sin (d x)) (e \sec (c+d x))^{5/2} (\cos (c+3 d x)+i \sin (c+3 d x)) \left (7 i e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-28 i \cos (2 (c+d x))+27 \tan (c+d x)+7 \sin (3 (c+d x)) \sec (c+d x)-36 i\right )}{70 d} \]

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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Maple [B]  time = 0.365, size = 365, normalized size = 3. \begin{align*}{\frac{2\,a \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{35\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( 21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -21\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -21\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+14\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+5\,i\sin \left ( dx+c \right ) +7\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{7}{2}}}} \end{align*}

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*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(21*I*cos(d*x+c)^4*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-21*I*cos(d*x+c)^4*sin(d*x+c)*(1/(cos(d*x+c)+1))
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+21*I*cos(d*x+c)^3*sin(d*x+c)
*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)-21*I*cos(
d*x+c)^3*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)