3.284 \(\int \sqrt {c e+d e x} (a+b \sin ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=99 \[ \frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}+\frac {4 b \sqrt {1-(c+d x)^2} \sqrt {e (c+d x)}}{9 d}-\frac {4 b \sqrt {e} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{9 d} \]

[Out]

2/3*(e*(d*x+c))^(3/2)*(a+b*arcsin(d*x+c))/d/e-4/9*b*EllipticF((e*(d*x+c))^(1/2)/e^(1/2),I)*e^(1/2)/d+4/9*b*(e*
(d*x+c))^(1/2)*(1-(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4805, 4627, 321, 329, 221} \[ \frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}+\frac {4 b \sqrt {1-(c+d x)^2} \sqrt {e (c+d x)}}{9 d}-\frac {4 b \sqrt {e} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*e + d*e*x]*(a + b*ArcSin[c + d*x]),x]

[Out]

(4*b*Sqrt[e*(c + d*x)]*Sqrt[1 - (c + d*x)^2])/(9*d) + (2*(e*(c + d*x))^(3/2)*(a + b*ArcSin[c + d*x]))/(3*d*e)
- (4*b*Sqrt[e]*EllipticF[ArcSin[Sqrt[e*(c + d*x)]/Sqrt[e]], -1])/(9*d)

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 321

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

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 \sqrt {c e+d e x} \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {e x} \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{3/2}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=\frac {4 b \sqrt {e (c+d x)} \sqrt {1-(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}-\frac {(2 b e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d}\\ &=\frac {4 b \sqrt {e (c+d x)} \sqrt {1-(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{9 d}\\ &=\frac {4 b \sqrt {e (c+d x)} \sqrt {1-(c+d x)^2}}{9 d}+\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e}-\frac {4 b \sqrt {e} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )\right |-1\right )}{9 d}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 87, normalized size = 0.88 \[ \frac {2 \sqrt {e (c+d x)} \left (3 a c+3 a d x-2 b \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right )+2 b \sqrt {1-(c+d x)^2}+3 b c \sin ^{-1}(c+d x)+3 b d x \sin ^{-1}(c+d x)\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*e + d*e*x]*(a + b*ArcSin[c + d*x]),x]

[Out]

(2*Sqrt[e*(c + d*x)]*(3*a*c + 3*a*d*x + 2*b*Sqrt[1 - (c + d*x)^2] + 3*b*c*ArcSin[c + d*x] + 3*b*d*x*ArcSin[c +
d*x] - 2*b*Hypergeometric2F1[1/4, 1/2, 5/4, (c + d*x)^2]))/(9*d)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d e x + c e} {\left (b \arcsin \left (d x + c\right ) + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(1/2)*(a+b*arcsin(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(1/2)*(a+b*arcsin(d*x+c)),x, algorithm="giac")

[Out]

integrate(sqrt(d*e*x + c*e)*(b*arcsin(d*x + c) + a), x)

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maple [B]  time = 0.01, size = 172, normalized size = 1.74 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {3}{2}} a}{3}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {3}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{3}-\frac {2 \left (-\frac {e^{2} \sqrt {d e x +c e}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{3}+\frac {e^{2} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )}{3 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{3 e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/e*(1/3*(d*e*x+c*e)^(3/2)*a+b*(1/3*(d*e*x+c*e)^(3/2)*arcsin((d*e*x+c*e)/e)-2/3/e*(-1/3*e^2*(d*e*x+c*e)^(1/2
)*(-(d*e*x+c*e)^2/e^2+1)^(1/2)+1/3*e^2/(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 \[ \frac {2 \, {\left (b d x + b c\right )} \sqrt {d x + c} \sqrt {e} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + 2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} a + d \int \frac {{\left (b d x + b c\right )} \sqrt {d x + c + 1} \sqrt {d x + c} \sqrt {-d x - c + 1}}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} \sqrt {e}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(1/2)*(a+b*arcsin(d*x+c)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^(1/2)*(a + b*asin(c + d*x)),x)

[Out]

int((c*e + d*e*x)^(1/2)*(a + b*asin(c + d*x)), x)

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sympy [A]  time = 2.40, size = 104, normalized size = 1.05 \[ \frac {2 a \left (c e + d e x\right )^{\frac {3}{2}}}{3 d e} + \frac {2 b \left (c e + d e x\right )^{\frac {3}{2}} \operatorname {asin}{\left (c + d x \right )}}{3 d e} - \frac {b \left (c e + d e x\right )^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {\left (c e + d e x\right )^{2} e^{2 i \pi }}{e^{2}}} \right )}}{3 d e^{2} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(1/2)*(a+b*asin(d*x+c)),x)

[Out]

2*a*(c*e + d*e*x)**(3/2)/(3*d*e) + 2*b*(c*e + d*e*x)**(3/2)*asin(c + d*x)/(3*d*e) - b*(c*e + d*e*x)**(5/2)*gam
ma(5/4)*hyper((1/2, 5/4), (9/4,), (c*e + d*e*x)**2*exp_polar(2*I*pi)/e**2)/(3*d*e**2*gamma(9/4))

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