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

Optimal. Leaf size=125 \[ \frac{10 a \sin (c+d x)}{21 d e^3 \sqrt{e \sec (c+d x)}}+\frac{10 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{21 d e^4}-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}} \]

[Out]

(((-2*I)/7)*a)/(d*(e*Sec[c + d*x])^(7/2)) + (10*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(21*d*e^4) + (2*a*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (10*a*Sin[c + d*x])/(21*d*e^3*Sqrt[e*S
ec[c + d*x]])

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Rubi [A]  time = 0.0823461, antiderivative size = 125, 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, 3769, 3771, 2641} \[ \frac{10 a \sin (c+d x)}{21 d e^3 \sqrt{e \sec (c+d x)}}+\frac{10 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{21 d e^4}-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/7)*a)/(d*(e*Sec[c + d*x])^(7/2)) + (10*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(21*d*e^4) + (2*a*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (10*a*Sin[c + d*x])/(21*d*e^3*Sqrt[e*S
ec[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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 2641

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

Rubi steps

\begin{align*} \int \frac{a+i a \tan (c+d x)}{(e \sec (c+d x))^{7/2}} \, dx &=-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+a \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx\\ &=-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac{(5 a) \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{7 e^2}\\ &=-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac{10 a \sin (c+d x)}{21 d e^3 \sqrt{e \sec (c+d x)}}+\frac{(5 a) \int \sqrt{e \sec (c+d x)} \, dx}{21 e^4}\\ &=-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac{10 a \sin (c+d x)}{21 d e^3 \sqrt{e \sec (c+d x)}}+\frac{\left (5 a \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 e^4}\\ &=-\frac{2 i a}{7 d (e \sec (c+d x))^{7/2}}+\frac{10 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{21 d e^4}+\frac{2 a \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}+\frac{10 a \sin (c+d x)}{21 d e^3 \sqrt{e \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.67352, size = 121, normalized size = 0.97 \[ \frac{a \sqrt{e \sec (c+d x)} (\cos (c+d x)+i \sin (c+d x)) \left (5 \sin (c+d x)+5 \sin (3 (c+d x))-14 i \cos (c+d x)+2 i \cos (3 (c+d x))+20 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (c+d x)-i \sin (c+d x))\right )}{42 d e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[e*Sec[c + d*x]]*(Cos[c + d*x] + I*Sin[c + d*x])*((-14*I)*Cos[c + d*x] + (2*I)*Cos[3*(c + d*x)] + 20*Sq
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[c + d*x] - I*Sin[c + d*x]) + 5*Sin[c + d*x] + 5*Sin[3*(c + d*x
)]))/(42*d*e^4)

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Maple [A]  time = 0.196, size = 187, normalized size = 1.5 \begin{align*}{\frac{2\,a}{21\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \left ( -3\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5\,i\cos \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 ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +5\,i\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 ) +5\,\cos \left ( dx+c \right ) \sin \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((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(7/2),x)

[Out]

2/21*a/d*(-3*I*cos(d*x+c)^4+5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic
F(I*(cos(d*x+c)-1)/sin(d*x+c),I)+3*cos(d*x+c)^3*sin(d*x+c)+5*I*(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)+5*cos(d*x+c)*sin(d*x+c))/cos(d*x+c)^4/(e/cos(d*x+c))^(7/2
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (84 \, d e^{4} e^{\left (2 i \, d x + 2 i \, c\right )}{\rm integral}\left (-\frac{5 i \, \sqrt{2} a \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 )}}{21 \, d e^{4}}, x\right ) + \sqrt{2}{\left (-3 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 19 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, a\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 )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{84 \, d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84*(84*d*e^4*e^(2*I*d*x + 2*I*c)*integral(-5/21*I*sqrt(2)*a*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x
- 1/2*I*c)/(d*e^4), x) + sqrt(2)*(-3*I*a*e^(6*I*d*x + 6*I*c) - 19*I*a*e^(4*I*d*x + 4*I*c) - 9*I*a*e^(2*I*d*x +
2*I*c) + 7*I*a)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(d*e^4)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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