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

Optimal. Leaf size=139 \[ -\frac{4 e^4 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{21 a d \sqrt{e \sin (c+d x)}}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a d}-\frac{2 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a d}+\frac{2 e (e \sin (c+d x))^{5/2}}{5 a d} \]

[Out]

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

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Rubi [A]  time = 0.279631, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3872, 2839, 2564, 30, 2568, 2569, 2642, 2641} \[ \frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a d}-\frac{2 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a d}-\frac{4 e^4 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 a d \sqrt{e \sin (c+d x)}}+\frac{2 e (e \sin (c+d x))^{5/2}}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2569

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Sin[e + f*x])^
n*(a*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m
, 2*n]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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

Mathematica [A]  time = 0.677895, size = 122, normalized size = 0.88 \[ \frac{e^3 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \sin (c+d x)} \left (40 \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+\sqrt{\sin (c+d x)} (25 \cos (c+d x)-42 \cos (2 (c+d x))+15 \cos (3 (c+d x))+42)\right )}{105 a d \sqrt{\sin (c+d x)} (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Cos[(c + d*x)/2]^2*Sec[c + d*x]*(40*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (42 + 25*Cos[c + d*x] - 42*Cos[
2*(c + d*x)] + 15*Cos[3*(c + d*x)])*Sqrt[Sin[c + d*x]])*Sqrt[e*Sin[c + d*x]])/(105*a*d*(1 + Sec[c + d*x])*Sqrt
[Sin[c + d*x]])

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Maple [A]  time = 1.288, size = 128, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{2\,e}{5\,a} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{e}^{4}}{21\,a\cos \left ( dx+c \right ) } \left ( 3\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ) -5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+2\,\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \sqrt{e \sin \left (d x + c\right )} \sin \left (d x + c\right )}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(e^3*cos(d*x + c)^2 - e^3)*sqrt(e*sin(d*x + c))*sin(d*x + c)/(a*sec(d*x + c) + a), x)

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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*sin(d*x+c))**(7/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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