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

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

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, 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 30

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

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.69, size = 122, normalized size = 0.88 \[ \frac {e^3 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (40 F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )+\sqrt {\sin (c+d x)} (25 \cos (c+d x)-42 \cos (2 (c+d x))+15 \cos (3 (c+d x))+42)\right )}{105 a d \sqrt {\sin (c+d x)} (\sec (c+d x)+1)} \]

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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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )}{a \sec \left (d x + c\right ) + a}, x\right ) \]

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

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)

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maple [A]  time = 3.28, size = 128, normalized size = 0.92 \[ \frac {\frac {2 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 a}+\frac {2 e^{4} \left (3 \left (\sin ^{5}\left (d x +c \right )\right )+\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-5 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{21 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

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*e/a*(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*sin(d*x+c)+2)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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