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

Optimal. Leaf size=189 \[ \frac {4 e^3}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {4 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {4 e \sqrt {e \sin (c+d x)}}{a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.59, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3872, 2875, 2873, 2567, 2642, 2641, 2564, 14, 2569} \[ \frac {4 e^3}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {4 e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {4 e \sqrt {e \sin (c+d x)}}{a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 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 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 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 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 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))^2} \, dx &=\int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{3/2}}{(-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}{(e \sin (c+d x))^{5/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{5/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{5/2}}+\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{5/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx}{a^2}\\ &=-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {\left (2 e^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 a^2}-\frac {\left (2 e^2\right ) \int \frac {\cos ^2(c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{a^2}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{e^2}}{x^{5/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}-\frac {\left (4 e^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 a^2}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{5/2}}-\frac {1}{e^2 \sqrt {x}}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}-\frac {\left (2 e^2 \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^3}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\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^2 d \sqrt {e \sin (c+d x)}}+\frac {4 e \sqrt {e \sin (c+d x)}}{a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}-\frac {\left (4 e^2 \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^3}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos (c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\frac {2 e^3 \cos ^3(c+d x)}{3 a^2 d (e \sin (c+d x))^{3/2}}-\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)}}{a^2 d \sqrt {e \sin (c+d x)}}+\frac {4 e \sqrt {e \sin (c+d x)}}{a^2 d}-\frac {4 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 119, normalized size = 0.63 \[ \frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (e \sin (c+d x))^{3/2} \left (\frac {24 F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )}{\sin ^{\frac {3}{2}}(c+d x)}+(10 \cos (c+d x)-\cos (2 (c+d x))+15) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*((15 + 10*Cos[c + d*x] - Cos[2*(c + d*x)])*Csc[c + d*x]*Sec[(c + d*x)/2]^
2 + (24*EllipticF[(-2*c + Pi - 2*d*x)/4, 2])/Sin[c + d*x]^(3/2))*(e*Sin[c + d*x])^(3/2))/(3*a^2*d*(1 + Sec[c +
d*x])^2)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \sin \left (d x + c\right )} e \sin \left (d x + c\right )}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

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

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

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 4.30, size = 153, normalized size = 0.81 \[ -\frac {2 e^{3} \left (3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{6}\left (d x +c \right )\right )+6 \left (\cos ^{5}\left (d x +c \right )\right )+4 \left (\cos ^{4}\left (d x +c \right )\right )-14 \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right )+8 \cos \left (d x +c \right )\right )}{3 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )-1\right ) d} \]

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))^2,x)

[Out]

-2/3/a^2/(e*sin(d*x+c))^(3/2)/cos(d*x+c)/(cos(d*x+c)^2-1)*e^3*(3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*
sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-cos(d*x+c)^6+6*cos(d*x+c)^5+4*cos(d*x+c)^4-14*co
s(d*x+c)^3-3*cos(d*x+c)^2+8*cos(d*x+c))/d

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

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))^2,x, algorithm="maxima")

[Out]

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

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

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))^2,x)

[Out]

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

[Out]

Timed out

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