3.2.22 \(\int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx\) [122]

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

[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 = 0.15, 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 = {3957, 2918, 2644, 30, 2649, 2721, 2720} \begin {gather*} -\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} \end {gather*}

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 2644

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 2649

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 2720

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2918

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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 \text {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 = 13.52, size = 69, normalized size = 0.68 \begin {gather*} -\frac {2 \left (-2 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+(-3+\cos (c+d x)) \sqrt {\sin (c+d x)}\right ) (e \sin (c+d x))^{3/2}}{3 a d \sin ^{\frac {3}{2}}(c+d x)} \end {gather*}

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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Maple [A]
time = 0.18, size = 112, normalized size = 1.10

method	result	size
default	\(\frac {2 e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(112\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.82, size = 86, normalized size = 0.84 \begin {gather*} -\frac {2 \, {\left (\sqrt {2} \sqrt {-i} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} \sqrt {i} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\cos \left (d x + c\right ) e^{\frac {3}{2}} - 3 \, e^{\frac {3}{2}}\right )} \sqrt {\sin \left (d x + c\right )}\right )}}{3 \, a d} \end {gather*}

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]

-2/3*(sqrt(2)*sqrt(-I)*e^(3/2)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*sqrt(I)*e^(3
/2)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (cos(d*x + c)*e^(3/2) - 3*e^(3/2))*sqrt(sin(d*x
+ c)))/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

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]

Integral((e*sin(c + d*x))**(3/2)/(sec(c + d*x) + 1), x)/a

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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