3.3.94 \(\int \sqrt {c e+d e x} (a+b \text {ArcSin}(c+d x))^2 \, dx\) [294]

Optimal. Leaf size=130 \[ \frac {2 (e (c+d x))^{3/2} (a+b \text {ArcSin}(c+d x))^2}{3 d e}-\frac {8 b (e (c+d x))^{5/2} (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )}{15 d e^2}+\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3} \]

[Out]

2/3*(e*(d*x+c))^(3/2)*(a+b*arcsin(d*x+c))^2/d/e-8/15*b*(e*(d*x+c))^(5/2)*(a+b*arcsin(d*x+c))*hypergeom([1/2, 5
/4],[9/4],(d*x+c)^2)/d/e^2+16/105*b^2*(e*(d*x+c))^(7/2)*hypergeom([1, 7/4, 7/4],[9/4, 11/4],(d*x+c)^2)/d/e^3

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Rubi [A]
time = 0.14, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4889, 4723, 4805} \begin {gather*} \frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3}-\frac {8 b (e (c+d x))^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right ) (a+b \text {ArcSin}(c+d x))}{15 d e^2}+\frac {2 (e (c+d x))^{3/2} (a+b \text {ArcSin}(c+d x))^2}{3 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(a + b*ArcSin[c + d*x])^2)/(3*d*e) - (8*b*(e*(c + d*x))^(5/2)*(a + b*ArcSin[c + d*x])*H
ypergeometric2F1[1/2, 5/4, 9/4, (c + d*x)^2])/(15*d*e^2) + (16*b^2*(e*(c + d*x))^(7/2)*HypergeometricPFQ[{1, 7
/4, 7/4}, {9/4, 11/4}, (c + d*x)^2])/(105*d*e^3)

Rule 4723

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_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d +
e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e,
f, m}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[m]

Rule 4889

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 )^2 \, dx &=\frac {\text {Subst}\left (\int \sqrt {e x} \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{3/2} \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e}\\ &=\frac {2 (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{3 d e}-\frac {8 b (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )}{15 d e^2}+\frac {16 b^2 (e (c+d x))^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )}{105 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 107, normalized size = 0.82 \begin {gather*} \frac {2 (e (c+d x))^{3/2} \left (7 (a+b \text {ArcSin}(c+d x)) \left (5 (a+b \text {ArcSin}(c+d x))-4 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};(c+d x)^2\right )\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};(c+d x)^2\right )\right )}{105 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(3/2)*(7*(a + b*ArcSin[c + d*x])*(5*(a + b*ArcSin[c + d*x]) - 4*b*(c + d*x)*Hypergeometric2F1
[1/2, 5/4, 9/4, (c + d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, (c + d*x)^2]))
/(105*d*e)

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \sqrt {d e x +c e}\, \left (a +b \arcsin \left (d x +c \right )\right )^{2}\, dx\]

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))^2,x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

1/6*(4*(b^2*d*x + b^2*c)*sqrt(d*x + c)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2*e^(1/2) + (36*
a*b*d^2*e^(1/2)*integrate(1/3*sqrt(d*x + c)*x^2*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*
x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 72*a*b*c*d*e^(1/2)*integrate(1/3*sqrt(d*x
+ c)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2
+ 2*c*d*x + c^2 - 1), x) + 36*a*b*c^2*e^(1/2)*integrate(1/3*sqrt(d*x + c)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-
d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 3*a^2*c^2*(2*arc
tan(sqrt(d*x + c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))*e^(1/2)/d + 24*b^2*d*e^(1/2)*integrate(1
/3*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/
(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 24*b^2*c*e^(1/2)*integrate(1/3*sqr
t(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*
x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) - 6*(2*(c + 1)*arctan(sqrt(d*x + c)) - (c -
1)*log(sqrt(d*x + c) + 1) + (c - 1)*log(sqrt(d*x + c) - 1) - 4*sqrt(d*x + c))*a^2*c*e^(1/2)/d - 36*a*b*e^(1/2)
*integrate(1/3*sqrt(d*x + c)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*
x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + (6*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c)) - 3*(c^2 - 2*c + 1)*
log(sqrt(d*x + c) + 1) + 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) - 1) + 4*(d*x + c)^(3/2) - 24*sqrt(d*x + c)*c)*a^
2*e^(1/2)/d - 3*a^2*(2*arctan(sqrt(d*x + c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))*e^(1/2)/d)*d)/
d

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {e \left (c + d x\right )} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}

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))**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {c\,e+d\,e\,x}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}

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))^2,x)

[Out]

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

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