\(\int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx\) [291]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {8 b (e (c+d x))^{11/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )}{99 d e^2}+\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3} \]

[Out]

2/9*(e*(d*x+c))^(9/2)*(a+b*arcsin(d*x+c))^2/d/e-8/99*b*(e*(d*x+c))^(11/2)*(a+b*arcsin(d*x+c))*hypergeom([1/2,
11/4],[15/4],(d*x+c)^2)/d/e^2+16/1287*b^2*(e*(d*x+c))^(13/2)*hypergeom([1, 13/4, 13/4],[15/4, 17/4],(d*x+c)^2)
/d/e^3

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4889, 4723, 4805} \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3}-\frac {8 b (e (c+d x))^{11/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right ) (a+b \arcsin (c+d x))}{99 d e^2}+\frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e} \]

[In]

Int[(c*e + d*e*x)^(7/2)*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(2*(e*(c + d*x))^(9/2)*(a + b*ArcSin[c + d*x])^2)/(9*d*e) - (8*b*(e*(c + d*x))^(11/2)*(a + b*ArcSin[c + d*x])*
Hypergeometric2F1[1/2, 11/4, 15/4, (c + d*x)^2])/(99*d*e^2) + (16*b^2*(e*(c + d*x))^(13/2)*HypergeometricPFQ[{
1, 13/4, 13/4}, {15/4, 17/4}, (c + d*x)^2])/(1287*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*} \text {integral}& = \frac {\text {Subst}\left (\int (e x)^{7/2} (a+b \arcsin (x))^2 \, dx,x,c+d x\right )}{d} \\ & = \frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {(4 b) \text {Subst}\left (\int \frac {(e x)^{9/2} (a+b \arcsin (x))}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{9 d e} \\ & = \frac {2 (e (c+d x))^{9/2} (a+b \arcsin (c+d x))^2}{9 d e}-\frac {8 b (e (c+d x))^{11/2} (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )}{99 d e^2}+\frac {16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )}{1287 d e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\frac {2 e^3 (c+d x)^4 \sqrt {e (c+d x)} \left (13 (a+b \arcsin (c+d x)) \left (11 (a+b \arcsin (c+d x))-4 b (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{4},\frac {15}{4},(c+d x)^2\right )\right )+8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {13}{4},\frac {13}{4};\frac {15}{4},\frac {17}{4};(c+d x)^2\right )\right )}{1287 d} \]

[In]

Integrate[(c*e + d*e*x)^(7/2)*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(2*e^3*(c + d*x)^4*Sqrt[e*(c + d*x)]*(13*(a + b*ArcSin[c + d*x])*(11*(a + b*ArcSin[c + d*x]) - 4*b*(c + d*x)*H
ypergeometric2F1[1/2, 11/4, 15/4, (c + d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 13/4, 13/4}, {15/4,
17/4}, (c + d*x)^2]))/(1287*d)

Maple [F]

\[\int \left (d e x +c e \right )^{\frac {7}{2}} \left (a +b \arcsin \left (d x +c \right )\right )^{2}d x\]

[In]

int((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c))^2,x)

[Out]

int((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c))^2,x)

Fricas [F]

\[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate((d*e*x+c*e)**(7/2)*(a+b*asin(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{\frac {7}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )}^{2} \,d x } \]

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)*(b*arcsin(d*x + c) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (c e+d e x)^{7/2} (a+b \arcsin (c+d x))^2 \, dx=\int {\left (c\,e+d\,e\,x\right )}^{7/2}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]

[In]

int((c*e + d*e*x)^(7/2)*(a + b*asin(c + d*x))^2,x)

[Out]

int((c*e + d*e*x)^(7/2)*(a + b*asin(c + d*x))^2, x)