3.1.87 \(\int \frac {(d-c^2 d x^2)^{5/2} (a+b \text {ArcSin}(c x))}{x^2} \, dx\) [87]

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

[Out]

-5/4*c^2*d*x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))-(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x-15/8*c^2*d^2*x*(a
+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)+9/16*b*c^3*d^2*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/16*b*c^5*d^2
*x^4*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-15/16*c*d^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/(-c^2*x^2+
1)^(1/2)+b*c*d^2*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 268, normalized size of antiderivative = 1.00, 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 = {4785, 4743, 4741, 4737, 30, 14, 272, 45} \begin {gather*} -\frac {15}{8} c^2 d^2 x \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))-\frac {15 c d^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^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} (a+b \text {ArcSin}(c x))-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \text {ArcSin}(c x))}{x}+\frac {b c d^2 \log (x) \sqrt {d-c^2 d x^2}}{\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 {9 b c^3 d^2 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}} \end {gather*}

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 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 30

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

Rule 45

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])

Rule 272

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 4737

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

Rule 4741

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[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcSin[c*x])^n/S
qrt[1 - c^2*x^2], x], x] - Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^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 4743

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcS
in[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]

Rule 4785

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/(f*(m + 1)))*Simp[(d + e*x^2)^p/(1 - c
^2*x^2)^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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.59, size = 257, normalized size = 0.96 \begin {gather*} \frac {d^2 \left (-120 b c x \sqrt {d-c^2 d x^2} \text {ArcSin}(c x)^2+240 a c \sqrt {d} x \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+\sqrt {d-c^2 d x^2} \left (-32 b c x \cos (2 \text {ArcSin}(c x))-b c x \cos (4 \text {ArcSin}(c x))+16 \left (a \sqrt {1-c^2 x^2} \left (-8-9 c^2 x^2+2 c^4 x^4\right )+8 b c x \log (c x)\right )\right )-4 b \sqrt {d-c^2 d x^2} \text {ArcSin}(c x) \left (32 \sqrt {1-c^2 x^2}+16 c x \sin (2 \text {ArcSin}(c x))+c x \sin (4 \text {ArcSin}(c x))\right )\right )}{128 x \sqrt {1-c^2 x^2}} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.25, size = 1391, normalized size = 5.19

method	result	size
default	\(\text {Expression too large to display}\)	\(1391\)

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,method=_RETURNVERBOSE)

[Out]

-17/64*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*x^2+31/256*b*(-d*(c^2*x^2-1
))^(1/2)*sin(3*arcsin(c*x))*d^2*c^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+79/64*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^
2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)*d^2*c-15/64*I*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)
*arcsin(c*x)-31/256*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*x^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))-a/d/x*(-c^2*d*x^2+d)^(7/2)-a*c^2*x*(-c^2*d*x^2+d)^(5/2)+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*
d^2/(c^2*x^2-1)/x+33/256*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+33/256*b*(-d*(c^2*x^2-1
))^(1/2)*cos(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)-15/64*b*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c^2/(c^2*x
^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x-1/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^5/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^
2+1)^(1/2)*x^4-3/8*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2+15/64*I*b
*(-d*(c^2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*arcsin(c*x)*x^2-33/256*I*b*(-d*(c^2*x^2-1))^(1/
2)*cos(3*arcsin(c*x))*d^2*c^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-17/64*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c
*x))*d^2*c^2/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x+1/32*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^5/(c^2*x^2-1)*(-
c^2*x^2+1)^(1/2)*x^4-9/32*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-33/256*b*(-d*(c^
2*x^2-1))^(1/2)*cos(3*arcsin(c*x))*d^2*c^3/(c^2*x^2-1)*x^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*d^2*c-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)*d^2*c+1/8*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^6/(c^2*x^2-1)*arcsin(c*x)*x^5-11/16*b*(-d*(c^2*x^2-1))^(1
/2)*d^2*c^4/(c^2*x^2-1)*arcsin(c*x)*x^3-7/16*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^2/(c^2*x^2-1)*arcsin(c*x)*x+17/64*
b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)*arcsin(c*x)+1/32*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*
c^6/(c^2*x^2-1)*x^5+13/64*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2*c^4/(c^2*x^2-1)*x^3-15/64*I*b*(-d*(c^2*x^2-1))^(1/2)*
d^2*c^2/(c^2*x^2-1)*x+31/256*I*b*(-d*(c^2*x^2-1))^(1/2)*sin(3*arcsin(c*x))*d^2*c/(c^2*x^2-1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

b*sqrt(d)*integrate((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1
)*sqrt(-c*x + 1))/x^2, x) - 1/8*(10*(-c^2*d*x^2 + d)^(3/2)*c^2*d*x + 15*sqrt(-c^2*d*x^2 + d)*c^2*d^2*x + 15*c*
d^(5/2)*arcsin(c*x) + 8*(-c^2*d*x^2 + d)^(5/2)/x)*a

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{2}}\, dx \end {gather*}

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]

Integral((-d*(c*x - 1)*(c*x + 1))**(5/2)*(a + b*asin(c*x))/x**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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