3.122 \(\int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=102 \[ -\frac {4 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 a d \sqrt {e \sin (c+d x)}}+\frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3872, 2839, 2564, 30, 2569, 2642, 2641} \[ -\frac {4 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 a d \sqrt {e \sin (c+d x)}}+\frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^(3/2)/(a + a*Sec[c + d*x]),x]

[Out]

(-4*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*a*d*Sqrt[e*Sin[c + d*x]]) + (2*e*Sqrt[e*Sin[c
+ d*x]])/(a*d) - (2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*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 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))^{3/2}}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) (e \sin (c+d x))^{3/2}}{-a-a \cos (c+d x)} \, dx\\ &=\frac {e^2 \int \frac {\cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a}-\frac {e^2 \int \frac {\cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a}\\ &=-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d}+\frac {e \operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,e \sin (c+d x)\right )}{a d}-\frac {\left (2 e^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 a}\\ &=\frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d}-\frac {\left (2 e^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 a \sqrt {e \sin (c+d x)}}\\ &=-\frac {4 e^2 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 a d \sqrt {e \sin (c+d x)}}+\frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d}\\ \end {align*}

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Mathematica [A]  time = 20.65, size = 69, normalized size = 0.68 \[ -\frac {2 (e \sin (c+d x))^{3/2} \left (\sqrt {\sin (c+d x)} (\cos (c+d x)-3)-2 F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{3 a d \sin ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sin[c + d*x])^(3/2)/(a + a*Sec[c + d*x]),x]

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \sin \left (d x + c\right )} e \sin \left (d x + c\right )}{a \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral(sqrt(e*sin(d*x + c))*e*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 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^(3/2)/(a*sec(d*x + c) + a), x)

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maple [A]  time = 3.28, size = 112, normalized size = 1.10 \[ \frac {2 e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

2/3/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip
ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-cos(d*x+c)^2*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^(3/2)/(a*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(3/2)/(a + a/cos(c + d*x)),x)

[Out]

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

[Out]

Timed out

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