3.87 \(\int \frac{(d-c^2 d x^2)^{5/2} (a+b \sin ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=268 \[ -\frac{15}{8} c^2 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{15 c d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt{1-c^2 x^2}}-\frac{5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{b c^5 d^2 x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{b c d^2 \log (x) \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

[Out]

(9*b*c^3*d^2*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1
- c^2*x^2]) - (15*c^2*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/8 - (5*c^2*d*x*(d - c^2*d*x^2)^(3/2)*(a +
b*ArcSin[c*x]))/4 - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x - (15*c*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcS
in[c*x])^2)/(16*b*Sqrt[1 - c^2*x^2]) + (b*c*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/Sqrt[1 - c^2*x^2]

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Rubi [A]  time = 0.239901, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {4695, 4649, 4647, 4641, 30, 14, 266, 43} \[ -\frac{15}{8} c^2 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{15 c d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt{1-c^2 x^2}}-\frac{5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{b c^5 d^2 x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{9 b c^3 d^2 x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{b c d^2 \log (x) \sqrt{d-c^2 d x^2}}{\sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b*c^3*d^2*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^4*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1
- c^2*x^2]) - (15*c^2*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/8 - (5*c^2*d*x*(d - c^2*d*x^2)^(3/2)*(a +
b*ArcSin[c*x]))/4 - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x - (15*c*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcS
in[c*x])^2)/(16*b*Sqrt[1 - c^2*x^2]) + (b*c*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/Sqrt[1 - c^2*x^2]

Rule 4695

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/
(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1),
x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 4649

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*(
a + b*ArcSin[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n,
x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 0]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\left (5 c^2 d\right ) \int \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx+\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \int \frac{\left (1-c^2 x^2\right )^2}{x} \, dx}{\sqrt{1-c^2 x^2}}\\ &=-\frac{5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{1}{4} \left (15 c^2 d^2\right ) \int \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx+\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-c^2 x\right )^2}{x} \, dx,x,x^2\right )}{2 \sqrt{1-c^2 x^2}}+\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \sqrt{1-c^2 x^2}}\\ &=-\frac{15}{8} c^2 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \left (-2 c^2+\frac{1}{x}+c^4 x\right ) \, dx,x,x^2\right )}{2 \sqrt{1-c^2 x^2}}-\frac{\left (15 c^2 d^2 \sqrt{d-c^2 d x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{8 \sqrt{1-c^2 x^2}}+\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \sqrt{1-c^2 x^2}}+\frac{\left (15 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt{1-c^2 x^2}}\\ &=\frac{9 b c^3 d^2 x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{15}{8} c^2 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{5}{4} c^2 d x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{15 c d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b \sqrt{1-c^2 x^2}}+\frac{b c d^2 \sqrt{d-c^2 d x^2} \log (x)}{\sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.887622, size = 257, normalized size = 0.96 \[ \frac{d^2 \left (\sqrt{d-c^2 d x^2} \left (16 \left (a \sqrt{1-c^2 x^2} \left (2 c^4 x^4-9 c^2 x^2-8\right )+8 b c x \log (c x)\right )-32 b c x \cos \left (2 \sin ^{-1}(c x)\right )-b c x \cos \left (4 \sin ^{-1}(c x)\right )\right )+240 a c \sqrt{d} x \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )-120 b c x \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)^2-4 b \sqrt{d-c^2 d x^2} \left (32 \sqrt{1-c^2 x^2}+16 c x \sin \left (2 \sin ^{-1}(c x)\right )+c x \sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)\right )}{128 x \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-120*b*c*x*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2 + 240*a*c*Sqrt[d]*x*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d -
c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + Sqrt[d - c^2*d*x^2]*(-32*b*c*x*Cos[2*ArcSin[c*x]] - b*c*x*Cos[4*ArcSin
[c*x]] + 16*(a*Sqrt[1 - c^2*x^2]*(-8 - 9*c^2*x^2 + 2*c^4*x^4) + 8*b*c*x*Log[c*x])) - 4*b*Sqrt[d - c^2*d*x^2]*A
rcSin[c*x]*(32*Sqrt[1 - c^2*x^2] + 16*c*x*Sin[2*ArcSin[c*x]] + c*x*Sin[4*ArcSin[c*x]])))/(128*x*Sqrt[1 - c^2*x
^2])

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Maple [C]  time = 0.25, size = 593, normalized size = 2.2 \begin{align*} -{\frac{a}{dx} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{7}{2}}}}-a{c}^{2}x \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{5}{2}}}-{\frac{5\,a{c}^{2}dx}{4} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{15\,a{c}^{2}{d}^{2}x}{8}\sqrt{-{c}^{2}d{x}^{2}+d}}-{\frac{15\,a{c}^{2}{d}^{3}}{8}\arctan \left ({x\sqrt{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{15\,b \left ( \arcsin \left ( cx \right ) \right ) ^{2}c{d}^{2}}{16\,{c}^{2}{x}^{2}-16}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{ib\arcsin \left ( cx \right ) c{d}^{2}}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b{d}^{2}\arcsin \left ( cx \right ) }{x \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bc{d}^{2}}{{c}^{2}{x}^{2}-1}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ) }+{\frac{b{c}^{6}{d}^{2}\arcsin \left ( cx \right ){x}^{5}}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{11\,b{c}^{4}{d}^{2}\arcsin \left ( cx \right ){x}^{3}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b{c}^{2}{d}^{2}\arcsin \left ( cx \right ) x}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{33\,bc{d}^{2}}{128\,{c}^{2}{x}^{2}-128}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b{c}^{5}{d}^{2}{x}^{4}}{16\,{c}^{2}{x}^{2}-16}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{9\,b{c}^{3}{d}^{2}{x}^{2}}{16\,{c}^{2}{x}^{2}-16}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d/x*(-c^2*d*x^2+d)^(7/2)-a*c^2*x*(-c^2*d*x^2+d)^(5/2)-5/4*a*c^2*d*x*(-c^2*d*x^2+d)^(3/2)-15/8*a*c^2*d^2*x*(
-c^2*d*x^2+d)^(1/2)-15/8*a*c^2*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+15/16*b*(-d*(c^2
*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)^2*c*d^2+I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2
)/(c^2*x^2-1)*arcsin(c*x)*c*d^2+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*d^2/(c^2*x^2-1)/x-b*(-d*(c^2*x^2-1))^(1/2
)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*c*d^2+1/4*b*(-d*(c^2*x^2-1))^(1/2)*c^6*d^2
/(c^2*x^2-1)*arcsin(c*x)*x^5-11/8*b*(-d*(c^2*x^2-1))^(1/2)*c^4*d^2/(c^2*x^2-1)*arcsin(c*x)*x^3+1/8*b*(-d*(c^2*
x^2-1))^(1/2)*c^2*d^2/(c^2*x^2-1)*arcsin(c*x)*x+33/128*b*(-d*(c^2*x^2-1))^(1/2)*c*d^2/(c^2*x^2-1)*(-c^2*x^2+1)
^(1/2)+1/16*b*(-d*(c^2*x^2-1))^(1/2)*c^5*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4-9/16*b*(-d*(c^2*x^2-1))^(1/2)*
c^3*d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqr
t(-c^2*d*x^2 + d)/x^2, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}{\left (b \arcsin \left (c x\right ) + a\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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