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

Optimal. Leaf size=162 \[ \frac{52 e^4 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{21 a^2 d \sqrt{e \sin (c+d x)}}-\frac{4 e^3 \sqrt{e \sin (c+d x)}}{a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}+\frac{26 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a^2 d}+\frac{4 e (e \sin (c+d x))^{5/2}}{5 a^2 d} \]

[Out]

(52*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a^2*d*Sqrt[e*Sin[c + d*x]]) - (4*e^3*Sqrt[e*S
in[c + d*x]])/(a^2*d) + (26*e^3*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(21*a^2*d) + (2*e^3*Cos[c + d*x]^3*Sqrt[e*S
in[c + d*x]])/(7*a^2*d) + (4*e*(e*Sin[c + d*x])^(5/2))/(5*a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.551093, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3872, 2875, 2873, 2569, 2642, 2641, 2564, 14} \[ -\frac{4 e^3 \sqrt{e \sin (c+d x)}}{a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}+\frac{26 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a^2 d}+\frac{52 e^4 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 a^2 d \sqrt{e \sin (c+d x)}}+\frac{4 e (e \sin (c+d x))^{5/2}}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(52*e^4*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*a^2*d*Sqrt[e*Sin[c + d*x]]) - (4*e^3*Sqrt[e*S
in[c + d*x]])/(a^2*d) + (26*e^3*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(21*a^2*d) + (2*e^3*Cos[c + d*x]^3*Sqrt[e*S
in[c + d*x]])/(7*a^2*d) + (4*e*(e*Sin[c + d*x])^(5/2))/(5*a^2*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 2875

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

Rule 2873

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

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{(e \sin (c+d x))^{7/2}}{(a+a \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) (e \sin (c+d x))^{7/2}}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac{e^4 \int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sqrt{e \sin (c+d x)}} \, dx}{a^4}\\ &=\frac{e^4 \int \left (\frac{a^2 \cos ^2(c+d x)}{\sqrt{e \sin (c+d x)}}-\frac{2 a^2 \cos ^3(c+d x)}{\sqrt{e \sin (c+d x)}}+\frac{a^2 \cos ^4(c+d x)}{\sqrt{e \sin (c+d x)}}\right ) \, dx}{a^4}\\ &=\frac{e^4 \int \frac{\cos ^2(c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{a^2}+\frac{e^4 \int \frac{\cos ^4(c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{a^2}-\frac{\left (2 e^4\right ) \int \frac{\cos ^3(c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{a^2}\\ &=\frac{2 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{1-\frac{x^2}{e^2}}{\sqrt{x}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}+\frac{\left (2 e^4\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{3 a^2}+\frac{\left (6 e^4\right ) \int \frac{\cos ^2(c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{7 a^2}\\ &=\frac{26 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{\sqrt{x}}-\frac{x^{3/2}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}+\frac{\left (4 e^4\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{7 a^2}+\frac{\left (2 e^4 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 a^2 \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)}}{3 a^2 d \sqrt{e \sin (c+d x)}}-\frac{4 e^3 \sqrt{e \sin (c+d x)}}{a^2 d}+\frac{26 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}+\frac{4 e (e \sin (c+d x))^{5/2}}{5 a^2 d}+\frac{\left (4 e^4 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{7 a^2 \sqrt{e \sin (c+d x)}}\\ &=\frac{52 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^2 d \sqrt{e \sin (c+d x)}}-\frac{4 e^3 \sqrt{e \sin (c+d x)}}{a^2 d}+\frac{26 e^3 \cos (c+d x) \sqrt{e \sin (c+d x)}}{21 a^2 d}+\frac{2 e^3 \cos ^3(c+d x) \sqrt{e \sin (c+d x)}}{7 a^2 d}+\frac{4 e (e \sin (c+d x))^{5/2}}{5 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.55736, size = 94, normalized size = 0.58 \[ -\frac{e^3 \sqrt{e \sin (c+d x)} \left (520 \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )+\sqrt{\sin (c+d x)} (-305 \cos (c+d x)+84 \cos (2 (c+d x))-15 \cos (3 (c+d x))+756)\right )}{210 a^2 d \sqrt{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*(520*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (756 - 305*Cos[c + d*x] + 84*Cos[2*(c + d*x)] - 15*Cos[3*(c +
d*x)])*Sqrt[Sin[c + d*x]])*Sqrt[e*Sin[c + d*x]])/(210*a^2*d*Sqrt[Sin[c + d*x]])

________________________________________________________________________________________

Maple [A]  time = 1.74, size = 145, normalized size = 0.9 \begin{align*} -{\frac{2\,{e}^{4}}{105\,{a}^{2}\cos \left ( dx+c \right ) d} \left ( -15\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+65\,\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},1/2\,\sqrt{2} \right ) +42\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -65\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +168\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \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))^2,x)

[Out]

-2/105/a^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^4*(-15*sin(d*x+c)*cos(d*x+c)^4+65*(-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))+42*cos(d*x+c)^3*sin(d*x+c)-65*cos(
d*x+c)^2*sin(d*x+c)+168*cos(d*x+c)*sin(d*x+c))/d

________________________________________________________________________________________

Maxima [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))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, 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))^2,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^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c
) + a^2), x)

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

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}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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