\(\int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx\) [297]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=-\frac {2 (a+b \arcsin (c+d x))^2}{3 d e (e (c+d x))^{3/2}}-\frac {8 b (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},(c+d x)^2\right )}{3 d e^2 \sqrt {e (c+d x)}}+\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3} \]

[Out]

-2/3*(a+b*arcsin(d*x+c))^2/d/e/(e*(d*x+c))^(3/2)-8/3*b*(a+b*arcsin(d*x+c))*hypergeom([-1/4, 1/2],[3/4],(d*x+c)
^2)/d/e^2/(e*(d*x+c))^(1/2)+16/3*b^2*hypergeom([1/4, 1/4, 1],[3/4, 5/4],(d*x+c)^2)*(e*(d*x+c))^(1/2)/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} \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3}-\frac {8 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},(c+d x)^2\right ) (a+b \arcsin (c+d x))}{3 d e^2 \sqrt {e (c+d x)}}-\frac {2 (a+b \arcsin (c+d x))^2}{3 d e (e (c+d x))^{3/2}} \]

[In]

Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(5/2),x]

[Out]

(-2*(a + b*ArcSin[c + d*x])^2)/(3*d*e*(e*(c + d*x))^(3/2)) - (8*b*(a + b*ArcSin[c + d*x])*Hypergeometric2F1[-1
/4, 1/2, 3/4, (c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) + (16*b^2*Sqrt[e*(c + d*x)]*HypergeometricPFQ[{1/4, 1/
4, 1}, {3/4, 5/4}, (c + d*x)^2])/(3*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 \frac {(a+b \arcsin (x))^2}{(e x)^{5/2}} \, dx,x,c+d x\right )}{d} \\ & = -\frac {2 (a+b \arcsin (c+d x))^2}{3 d e (e (c+d x))^{3/2}}+\frac {(4 b) \text {Subst}\left (\int \frac {a+b \arcsin (x)}{(e x)^{3/2} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d e} \\ & = -\frac {2 (a+b \arcsin (c+d x))^2}{3 d e (e (c+d x))^{3/2}}-\frac {8 b (a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},(c+d x)^2\right )}{3 d e^2 \sqrt {e (c+d x)}}+\frac {16 b^2 \sqrt {e (c+d x)} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )}{3 d e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=-\frac {2 \left ((a+b \arcsin (c+d x))^2+4 b (c+d x) \left ((a+b \arcsin (c+d x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},(c+d x)^2\right )-2 b (c+d x) \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};(c+d x)^2\right )\right )\right )}{3 d e (e (c+d x))^{3/2}} \]

[In]

Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {5}{2}}}d x\]

[In]

int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x)

[Out]

int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x)

Fricas [F]

\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}}{\left (e \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+b*asin(d*x+c))**2/(d*e*x+c*e)**(5/2),x)

[Out]

Integral((a + b*asin(c + d*x))**2/(e*(c + d*x))**(5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),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]

\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^2/(d*e*x + c*e)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{5/2}} \,d x \]

[In]

int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(5/2),x)

[Out]

int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^(5/2), x)