3.287 \(\int \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ -\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {4 b \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {c+d x}}-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}} \]

[Out]

-2/3*(a+b*arcsin(d*x+c))/d/e/(e*(d*x+c))^(3/2)+4/3*b*EllipticE(1/2*(-d*x-c+1)^(1/2)*2^(1/2),2^(1/2))*(e*(d*x+c
))^(1/2)/d/e^3/(d*x+c)^(1/2)-4/3*b*(1-(d*x+c)^2)^(1/2)/d/e^2/(e*(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4805, 4627, 325, 320, 318, 424} \[ -\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}}+\frac {4 b \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[1 - (c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) - (2*(a + b*ArcSin[c + d*x]))/(3*d*e*(e*(c + d*x))^(3
/2)) + (4*b*Sqrt[e*(c + d*x)]*EllipticE[ArcSin[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(3*d*e^3*Sqrt[c + d*x])

Rule 318

Int[Sqrt[x_]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[-2/(Sqrt[a]*(-(b/a))^(3/4)), Subst[Int[Sqrt[1 - 2*x^
2]/Sqrt[1 - x^2], x], x, Sqrt[1 - Sqrt[-(b/a)]*x]/Sqrt[2]], x] /; FreeQ[{a, b}, x] && GtQ[-(b/a), 0] && GtQ[a,
0]

Rule 320

Int[Sqrt[(c_)*(x_)]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[c*x]/Sqrt[x], Int[Sqrt[x]/Sqrt[a + b*x^2
], x], x] /; FreeQ[{a, b, c}, x] && GtQ[-(b/a), 0]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{(e x)^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{(e x)^{3/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^3}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac {\left (2 b \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e^3 \sqrt {c+d x}}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {\left (4 b \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-2 x^2}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )}{3 d e^3 \sqrt {c+d x}}\\ &=-\frac {4 b \sqrt {1-(c+d x)^2}}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}+\frac {4 b \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )\right |2\right )}{3 d e^3 \sqrt {c+d x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 56, normalized size = 0.46 \[ -\frac {2 \left (a+2 b (c+d x) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};(c+d x)^2\right )+b \sin ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c + d*x] + 2*b*(c + d*x)*Hypergeometric2F1[-1/4, 1/2, 3/4, (c + d*x)^2]))/(3*d*e*(e*(c + d*x
))^(3/2))

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \arcsin \left (d x + c\right ) + a\right )}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*e*x + c*e)*(b*arcsin(d*x + c) + a)/(d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^3),
x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(d*x + c) + a)/(d*e*x + c*e)^(5/2), x)

________________________________________________________________________________________

maple [C]  time = 0.02, size = 190, normalized size = 1.56 \[ \frac {-\frac {2 a}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arcsin \left (\frac {d e x +c e}{e}\right )}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {-\frac {2 \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{3 \sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{3 e \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(5/2),x)

[Out]

2/d/e*(-1/3*a/(d*e*x+c*e)^(3/2)+b*(-1/3/(d*e*x+c*e)^(3/2)*arcsin((d*e*x+c*e)/e)+2/3/e*(-(-(d*e*x+c*e)^2/e^2+1)
^(1/2)/(d*e*x+c*e)^(1/2)+1/e/(1/e)^(1/2)*(1-(d*e*x+c*e)/e)^(1/2)*((d*e*x+c*e)/e+1)^(1/2)/(-(d*e*x+c*e)^2/e^2+1
)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(1/e)^(1/2),I)-EllipticE((d*e*x+c*e)^(1/2)*(1/e)^(1/2),I)))))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + {\left ({\left (b d^{2} e^{3} x + b c d e^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}} \int \frac {\sqrt {d x + c + 1} \sqrt {d x + c} \sqrt {-d x - c + 1}}{d^{4} e^{3} x^{4} + 4 \, c d^{3} e^{3} x^{3} + {\left (6 \, c^{2} - 1\right )} d^{2} e^{3} x^{2} + 2 \, {\left (2 \, c^{3} - c\right )} d e^{3} x + {\left (c^{4} - c^{2}\right )} e^{3}}\,{d x} + a d x + a c\right )} \sqrt {d x + c}\right )}}{3 \, {\left (d^{2} e^{2} x + c d e^{2}\right )} {\left (d x + c\right )}^{2} \sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^(5/2),x, algorithm="maxima")

[Out]

-2/3*((d*x + c)^(3/2)*b*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + (3*(b*d^2*e^3*x + b*c*d*e^3)*
(d*x + c)^(3/2)*integrate(1/3*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)/(d^4*e^3*x^4 + 4*c*d^3*e^3*x^
3 + (6*c^2 - 1)*d^2*e^3*x^2 + 2*(2*c^3 - c)*d*e^3*x + (c^4 - c^2)*e^3), x) + a*d*x + a*c)*sqrt(d*x + c))/((d^2
*e^2*x + c*d*e^2)*(d*x + c)^2*sqrt(e))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(5/2),x)

[Out]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))/(d*e*x+c*e)**(5/2),x)

[Out]

Integral((a + b*asin(c + d*x))/(e*(c + d*x))**(5/2), x)

________________________________________________________________________________________