\(\int \frac {(a+b \arcsin (c x))^2}{(d x)^{5/2}} \, dx\) [214]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int \frac {(a+b \arcsin (c x))^2}{(d x)^{5/2}} \, dx=-\frac {2 (a+b \arcsin (c x))^2}{3 d (d x)^{3/2}}-\frac {8 b c (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )}{3 d^2 \sqrt {d x}}+\frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4723, 4805} \[ \int \frac {(a+b \arcsin (c x))^2}{(d x)^{5/2}} \, dx=\frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3}-\frac {8 b c \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right ) (a+b \arcsin (c x))}{3 d^2 \sqrt {d x}}-\frac {2 (a+b \arcsin (c x))^2}{3 d (d x)^{3/2}} \]

[In]

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

[Out]

(-2*(a + b*ArcSin[c*x])^2)/(3*d*(d*x)^(3/2)) - (8*b*c*(a + b*ArcSin[c*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, c^
2*x^2])/(3*d^2*Sqrt[d*x]) + (16*b^2*c^2*Sqrt[d*x]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2])/(3*d^
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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (a+b \arcsin (c x))^2}{3 d (d x)^{3/2}}+\frac {(4 b c) \int \frac {a+b \arcsin (c x)}{(d x)^{3/2} \sqrt {1-c^2 x^2}} \, dx}{3 d} \\ & = -\frac {2 (a+b \arcsin (c x))^2}{3 d (d x)^{3/2}}-\frac {8 b c (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )}{3 d^2 \sqrt {d x}}+\frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b \arcsin (c x))^2}{(d x)^{5/2}} \, dx=\frac {x \left (-2 (a+b \arcsin (c x)) \left (a+b \arcsin (c x)+4 b c x \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )\right )+16 b^2 c^2 x^2 \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )\right )}{3 (d x)^{5/2}} \]

[In]

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

[Out]

(x*(-2*(a + b*ArcSin[c*x])*(a + b*ArcSin[c*x] + 4*b*c*x*Hypergeometric2F1[-1/4, 1/2, 3/4, c^2*x^2]) + 16*b^2*c
^2*x^2*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2]))/(3*(d*x)^(5/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

Maxima [F]

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

[In]

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

[Out]

-1/6*((3*a^2*c^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/(sqrt(c)*d^3) - log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqr
t(c)))/(sqrt(c)*d^3)) - 36*a*b*c^2*sqrt(d)*integrate(1/3*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c
^2*d^3*x^5 - d^3*x^3), x) + 24*b^2*c*sqrt(d)*integrate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(c*x/(sq
rt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*d^3*x^5 - d^3*x^3), x) - a^2*sqrt(d)*(6*c^(3/2)*arctan(sqrt(c)*sqrt(x))/d^3
- 3*c^(3/2)*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/d^3 - 4/(d^3*x^(3/2))) + 36*a*b*sqrt(d)*integrate
(1/3*sqrt(x)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*d^3*x^5 - d^3*x^3), x))*d^(5/2)*x^(3/2) + 4*b^2*a
rctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/(d^(5/2)*x^(3/2))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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