3.286 \(\int \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^{3/2}} \, dx\)

Optimal. Leaf size=61 \[ \frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{d e^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4805, 4627, 329, 221} \[ \frac {4 b F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{d e^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.03, size = 54, normalized size = 0.89 \[ -\frac {2 \left (a-2 b (c+d x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right )+b \sin ^{-1}(c+d x)\right )}{d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c + d*x] - 2*b*(c + d*x)*Hypergeometric2F1[1/4, 1/2, 5/4, (c + d*x)^2]))/(d*e*Sqrt[e*(c + d*
x)])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 132, normalized size = 2.16 \[ \frac {-\frac {2 a}{\sqrt {d e x +c e}}+2 b \left (-\frac {\arcsin \left (\frac {d e x +c e}{e}\right )}{\sqrt {d e x +c e}}+\frac {2 \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )}{e \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(d*x + c)*b*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + (sqrt(d*x + c)*b*d*e^2*in
tegrate(sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)/(d^3*e^2*x^3 + 3*c*d^2*e^2*x^2 + (3*c^2 - 1)*d*e^2*
x + (c^3 - c)*e^2), x) + a)*sqrt(d*x + c)*sqrt(e))/((d*x + c)*d*e^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asin}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(3/2),x)

[Out]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))/(d*e*x+c*e)**(3/2),x)

[Out]

Integral((a + b*asin(c + d*x))/(e*(c + d*x))**(3/2), x)

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