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

Optimal. Leaf size=117 \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}+\frac {4 b \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}}{25 d}+\frac {12 b e \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {c+d x}} \]

[Out]

2/5*(e*(d*x+c))^(5/2)*(a+b*arcsin(d*x+c))/d/e+12/25*b*e*EllipticE(1/2*(-d*x-c+1)^(1/2)*2^(1/2),2^(1/2))*(e*(d*
x+c))^(1/2)/d/(d*x+c)^(1/2)+4/25*b*(e*(d*x+c))^(3/2)*(1-(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.10, antiderivative size = 117, 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, 321, 320, 318, 424} \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}+\frac {4 b \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}}{25 d}+\frac {12 b e \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(e*(c + d*x))^(3/2)*Sqrt[1 - (c + d*x)^2])/(25*d) + (2*(e*(c + d*x))^(5/2)*(a + b*ArcSin[c + d*x]))/(5*d*
e) + (12*b*e*Sqrt[e*(c + d*x)]*EllipticE[ArcSin[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(25*d*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 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 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 (c e+d e x)^{3/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{3/2} \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=\frac {4 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}-\frac {(6 b e) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{25 d}\\ &=\frac {4 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}-\frac {\left (6 b e \sqrt {e (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{25 d \sqrt {c+d x}}\\ &=\frac {4 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}+\frac {\left (12 b e \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 )}{25 d \sqrt {c+d x}}\\ &=\frac {4 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )}{5 d e}+\frac {12 b e \sqrt {e (c+d x)} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {c+d x}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 87, normalized size = 0.74 \[ \frac {2 (e (c+d x))^{3/2} \left (5 a c+5 a d x-2 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )+2 b \sqrt {1-(c+d x)^2}+5 b c \sin ^{-1}(c+d x)+5 b d x \sin ^{-1}(c+d x)\right )}{25 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(5*a*c + 5*a*d*x + 2*b*Sqrt[1 - (c + d*x)^2] + 5*b*c*ArcSin[c + d*x] + 5*b*d*x*ArcSin[c
+ d*x] - 2*b*Hypergeometric2F1[1/2, 3/4, 7/4, (c + d*x)^2]))/(25*d)

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d e x + a c e + {\left (b d e x + b c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 194, normalized size = 1.66 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (-\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 e^{3} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {\frac {d e x +c e}{e}+1}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{5 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/e*(1/5*(d*e*x+c*e)^(5/2)*a+b*(1/5*(d*e*x+c*e)^(5/2)*arcsin((d*e*x+c*e)/e)-2/5/e*(-1/5*e^2*(d*e*x+c*e)^(3/2
)*(-(d*e*x+c*e)^2/e^2+1)^(1/2)-3/5*e^3/(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)))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (b d^{2} e x^{2} + 2 \, b c d e x + b c^{2} e\right )} \sqrt {d x + c} \sqrt {e} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + 2 \, {\left ({\left (d x + c\right )}^{\frac {5}{2}} a e + d \int \frac {{\left (b d^{2} e x^{2} + 2 \, b c d e x + b c^{2} e\right )} \sqrt {d x + c + 1} \sqrt {d x + c} \sqrt {-d x - c + 1}}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} \sqrt {e}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*(b*d^2*e*x^2 + 2*b*c*d*e*x + b*c^2*e)*sqrt(d*x + c)*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*
x - c + 1)) + (2*(d*x + c)^(5/2)*a*e + 5*d*integrate(2/5*(b*d^2*e*x^2 + 2*b*c*d*e*x + b*c^2*e)*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 \[ \int {\left (c\,e+d\,e\,x\right )}^{3/2}\,\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 19.22, size = 156, normalized size = 1.33 \[ a c e \left (\begin {cases} x \sqrt {c e} & \text {for}\: d = 0 \\0 & \text {for}\: e = 0 \\\frac {2 \left (c e + d e x\right )^{\frac {3}{2}}}{3 d e} & \text {otherwise} \end {cases}\right ) - \frac {2 a c \left (c e + d e x\right )^{\frac {3}{2}}}{3 d} + \frac {2 a \left (c e + d e x\right )^{\frac {5}{2}}}{5 d e} + \frac {2 b \left (c e + d e x\right )^{\frac {5}{2}} \operatorname {asin}{\left (c + d x \right )}}{5 d e} - \frac {b \left (c e + d e x\right )^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {\left (c e + d e x\right )^{2} e^{2 i \pi }}{e^{2}}} \right )}}{5 d e^{2} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c*e*Piecewise((x*sqrt(c*e), Eq(d, 0)), (0, Eq(e, 0)), (2*(c*e + d*e*x)**(3/2)/(3*d*e), True)) - 2*a*c*(c*e +
d*e*x)**(3/2)/(3*d) + 2*a*(c*e + d*e*x)**(5/2)/(5*d*e) + 2*b*(c*e + d*e*x)**(5/2)*asin(c + d*x)/(5*d*e) - b*(
c*e + d*e*x)**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), (c*e + d*e*x)**2*exp_polar(2*I*pi)/e**2)/(5*d*e**2*g
amma(11/4))

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