3.88 \(\int \frac{(d-c^2 d x^2)^{5/2} (a+b \sin ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=277 \[ \frac{5}{2} c^4 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^3 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt{1-c^2 x^2}}+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{b c^5 d^2 x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{1-c^2 x^2}}-\frac{7 b c^3 d^2 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{1-c^2 x^2}} \]

[Out]

-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(6*x^2*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^2*Sqrt[d - c^2*d*x^2])/(4*Sqrt[1 - c^2
*x^2]) + (5*c^4*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/2 + (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSi
n[c*x]))/(3*x) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(3*x^3) + (5*c^3*d^2*Sqrt[d - c^2*d*x^2]*(a + b*A
rcSin[c*x])^2)/(4*b*Sqrt[1 - c^2*x^2]) - (7*b*c^3*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/(3*Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.304384, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {4695, 4647, 4641, 30, 14, 266, 43} \[ \frac{5}{2} c^4 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^3 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt{1-c^2 x^2}}+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{b c^5 d^2 x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{1-c^2 x^2}}-\frac{7 b c^3 d^2 \log (x) \sqrt{d-c^2 d x^2}}{3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(6*x^2*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^2*Sqrt[d - c^2*d*x^2])/(4*Sqrt[1 - c^2
*x^2]) + (5*c^4*d^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/2 + (5*c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSi
n[c*x]))/(3*x) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(3*x^3) + (5*c^3*d^2*Sqrt[d - c^2*d*x^2]*(a + b*A
rcSin[c*x])^2)/(4*b*Sqrt[1 - c^2*x^2]) - (7*b*c^3*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/(3*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 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^4} \, dx &=-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{1}{3} \left (5 c^2 d\right ) \int \frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, 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^3} \, dx}{3 \sqrt{1-c^2 x^2}}\\ &=\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\left (5 c^4 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^2} \, dx,x,x^2\right )}{6 \sqrt{1-c^2 x^2}}-\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int \frac{1-c^2 x^2}{x} \, dx}{3 \sqrt{1-c^2 x^2}}\\ &=\frac{5}{2} c^4 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{\left (b c d^2 \sqrt{d-c^2 d x^2}\right ) \operatorname{Subst}\left (\int \left (c^4+\frac{1}{x^2}-\frac{2 c^2}{x}\right ) \, dx,x,x^2\right )}{6 \sqrt{1-c^2 x^2}}-\frac{\left (5 b c^3 d^2 \sqrt{d-c^2 d x^2}\right ) \int \left (\frac{1}{x}-c^2 x\right ) \, dx}{3 \sqrt{1-c^2 x^2}}+\frac{\left (5 c^4 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}{2 \sqrt{1-c^2 x^2}}-\frac{\left (5 b c^5 d^2 \sqrt{d-c^2 d x^2}\right ) \int x \, dx}{2 \sqrt{1-c^2 x^2}}\\ &=-\frac{b c d^2 \sqrt{d-c^2 d x^2}}{6 x^2 \sqrt{1-c^2 x^2}}-\frac{b c^5 d^2 x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}+\frac{5}{2} c^4 d^2 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{5 c^3 d^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt{1-c^2 x^2}}-\frac{7 b c^3 d^2 \sqrt{d-c^2 d x^2} \log (x)}{3 \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.51381, size = 243, normalized size = 0.88 \[ \frac{1}{24} d^2 \left (\frac{\sqrt{d-c^2 d x^2} \left (4 a \sqrt{1-c^2 x^2} \left (3 c^4 x^4+14 c^2 x^2-2\right )+b \left (-6 c^5 x^5+3 c^3 x^3-4 c x\right )-56 b c^3 x^3 \log (c x)\right )}{x^3 \sqrt{1-c^2 x^2}}-60 a c^3 \sqrt{d} \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )+\frac{30 b c^3 \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)^2}{\sqrt{1-c^2 x^2}}+\frac{4 b \left (3 c^4 x^4+14 c^2 x^2-2\right ) \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

(d^2*((4*b*Sqrt[d - c^2*d*x^2]*(-2 + 14*c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x])/x^3 + (30*b*c^3*Sqrt[d - c^2*d*x^2]*
ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 60*a*c^3*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))]
+ (Sqrt[d - c^2*d*x^2]*(4*a*Sqrt[1 - c^2*x^2]*(-2 + 14*c^2*x^2 + 3*c^4*x^4) + b*(-4*c*x + 3*c^3*x^3 - 6*c^5*x
^5) - 56*b*c^3*x^3*Log[c*x]))/(x^3*Sqrt[1 - c^2*x^2])))/24

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Maple [C]  time = 0.295, size = 1527, normalized size = 5.5 \begin{align*} \text{result too large to display} \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^4,x)

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(7/2)+4/3*a*c^4*x*(-c^2*d*x^2+d)^(5/2)+4/3*a*c^2/d/x*(-c^2*d*x^2+d)^(7/2)-1/8*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)-49/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15
*c^2*x^2+1)*x^5/(c^2*x^2-1)*c^8-203*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^3/(c^2*x^2-1)*arc
sin(c*x)*c^6+21/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^
5+190/3*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*arcsin(c*x)*c^4-23/3*b*(-d*(c^2*x
^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/x/(c^2*x^2-1)*arcsin(c*x)*c^2+1/6*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*
c^4*x^4-15*c^2*x^2+1)/x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+147*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2
*x^2+1)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^8+28/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^3/(c^2
*x^2-1)*c^6-7/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*c^4-14*I*b*(-c^2*x^2+1)
^(1/2)*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*c^3*d^2/(3*c^2*x^2-3)+1/4*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^2-5/4*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^3*d^2+5/
3*a*c^4*d*x*(-c^2*d*x^2+d)^(3/2)+5/2*a*c^4*d^2*x*(-c^2*d*x^2+d)^(1/2)+5/2*a*c^4*d^3/(c^2*d)^(1/2)*arctan((c^2*
d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+7/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/(c^2*x^2-1)*arcs
in(c*x)*(-c^2*x^2+1)^(1/2)*c^3-49/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^3/(c^2*x^2-1)*(
-c^2*x^2+1)*c^6+7/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x/(c^2*x^2-1)*(-c^2*x^2+1)*c^4+1/
2*b*(-d*(c^2*x^2-1))^(1/2)*c^6*d^2/(c^2*x^2-1)*arcsin(c*x)*x^3-1/2*b*(-d*(c^2*x^2-1))^(1/2)*c^4*d^2/(c^2*x^2-1
)*arcsin(c*x)*x-5/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)+
7/3*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^3*d^2+1/3*b*(
-d*(c^2*x^2-1))^(1/2)*d^2/(63*c^4*x^4-15*c^2*x^2+1)/x^3/(c^2*x^2-1)*arcsin(c*x)+147*I*b*(-d*(c^2*x^2-1))^(1/2)
*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^4/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^7-35*I*b*(-d*(c^2*x^2-1))^(1/2
)*d^2/(63*c^4*x^4-15*c^2*x^2+1)*x^2/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5

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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^4,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^{4}}, 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^4,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^4, 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**4,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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((-c^2*d*x^2 + d)^(5/2)*(b*arcsin(c*x) + a)/x^4, x)