3.128 \(\int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=187 \[ \frac {4 e^3}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {44 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d} \]

[Out]

4/3*e*(e*sin(d*x+c))^(3/2)/a^2/d-12/5*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/a^2/d+4*e^3/a^2/d/(e*sin(d*x+c))^(1/2)
-2*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(1/2)-2*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(1/2)+44/5*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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c)
)^(1/2)/a^2/d/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.60, antiderivative size = 187, 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, 2640, 2639, 2564, 14, 2569} \[ \frac {4 e^3}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {44 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*e^3)/(a^2*d*Sqrt[e*Sin[c + d*x]]) - (2*e^3*Cos[c + d*x])/(a^2*d*Sqrt[e*Sin[c + d*x]]) - (2*e^3*Cos[c + d*x]
^3)/(a^2*d*Sqrt[e*Sin[c + d*x]]) - (44*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*a^2*d*Sqr
t[Sin[c + d*x]]) + (4*e*(e*Sin[c + d*x])^(3/2))/(3*a^2*d) - (12*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(5*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 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[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))^{5/2}}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) (e \sin (c+d x))^{5/2}}{(-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))^{3/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{3/2}}+\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{3/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{a^2}\\ &=-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {\left (2 e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{a^2}-\frac {\left (6 e^2\right ) \int \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^{3/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d}-\frac {\left (12 e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{5 a^2}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/2}}-\frac {\sqrt {x}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}-\frac {\left (2 e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{a^2 \sqrt {\sin (c+d x)}}\\ &=\frac {4 e^3}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {4 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a^2 d \sqrt {\sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d}-\frac {\left (12 e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 a^2 \sqrt {\sin (c+d x)}}\\ &=\frac {4 e^3}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {44 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 2.99, size = 249, normalized size = 1.33 \[ \frac {4 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (e \sin (c+d x))^{5/2} \left (\csc ^2(c+d x) \left (20 \sin (c) \cos (d x)-3 \sin (2 c) \cos (2 d x)+20 \cos (c) \sin (d x)-3 \cos (2 c) \sin (2 d x)+\sec \left (\frac {c}{2}\right ) \left (60 \sin \left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )-36 \sin \left (\frac {3 c}{2}\right ) \sec (c)\right )-96 \tan \left (\frac {c}{2}\right ) \sec (c)\right )+\frac {352 i e^{2 i (2 c+d x)} \left (3 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};e^{2 i (c+d x)}\right )+e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right )\right )}{\left (1+e^{2 i c}\right ) \left (1-e^{2 i (c+d x)}\right )^{5/2}}\right )}{15 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sin[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(e*Sin[c + d*x])^(5/2)*(((352*I)*E^((2*I)*(2*c + d*x))*(3*Hypergeometric2
F1[-1/4, 1/2, 3/4, E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])
)/((1 + E^((2*I)*c))*(1 - E^((2*I)*(c + d*x)))^(5/2)) + Csc[c + d*x]^2*(20*Cos[d*x]*Sin[c] - 3*Cos[2*d*x]*Sin[
2*c] + Sec[c/2]*(-36*Sec[c]*Sin[(3*c)/2] + 60*Sec[(c + d*x)/2]*Sin[(d*x)/2]) + 20*Cos[c]*Sin[d*x] - 3*Cos[2*c]
*Sin[2*d*x] - 96*Sec[c]*Tan[c/2])))/(15*a^2*d*(1 + Sec[c + d*x])^2)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (e^{2} \cos \left (d x + c\right )^{2} - e^{2}\right )} \sqrt {e \sin \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-(e^2*cos(d*x + c)^2 - e^2)*sqrt(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 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^2, x)

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maple [A]  time = 4.03, size = 173, normalized size = 0.93 \[ -\frac {2 e^{3} \left (33 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-66 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \left (\cos ^{4}\left (d x +c \right )\right )+10 \left (\cos ^{3}\left (d x +c \right )\right )+33 \left (\cos ^{2}\left (d x +c \right )\right )-40 \cos \left (d x +c \right )\right )}{15 a^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-2/15/a^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^3*(33*(-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))-66*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2
)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*cos(d*x+c)^4+10*cos(d*x+c)^3+33*cos(d*x+c)^2-40*cos(d*x+c))/d

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maxima [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))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(5/2)/(a + a/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e*sin(c + d*x))^(5/2))/(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))**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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