3.79 \(\int x^3 (d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=227 \[ \frac{\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4 d^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d}+\frac{b c^3 d x^7 \sqrt{d-c^2 d x^2}}{49 \sqrt{1-c^2 x^2}}-\frac{8 b c d x^5 \sqrt{d-c^2 d x^2}}{175 \sqrt{1-c^2 x^2}}+\frac{b d x^3 \sqrt{d-c^2 d x^2}}{105 c \sqrt{1-c^2 x^2}}+\frac{2 b d x \sqrt{d-c^2 d x^2}}{35 c^3 \sqrt{1-c^2 x^2}} \]

[Out]

(2*b*d*x*Sqrt[d - c^2*d*x^2])/(35*c^3*Sqrt[1 - c^2*x^2]) + (b*d*x^3*Sqrt[d - c^2*d*x^2])/(105*c*Sqrt[1 - c^2*x
^2]) - (8*b*c*d*x^5*Sqrt[d - c^2*d*x^2])/(175*Sqrt[1 - c^2*x^2]) + (b*c^3*d*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqrt[
1 - c^2*x^2]) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d) + ((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c
*x]))/(7*c^4*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.199504, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {266, 43, 4691, 12, 373} \[ \frac{\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4 d^2}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d}+\frac{b c^3 d x^7 \sqrt{d-c^2 d x^2}}{49 \sqrt{1-c^2 x^2}}-\frac{8 b c d x^5 \sqrt{d-c^2 d x^2}}{175 \sqrt{1-c^2 x^2}}+\frac{b d x^3 \sqrt{d-c^2 d x^2}}{105 c \sqrt{1-c^2 x^2}}+\frac{2 b d x \sqrt{d-c^2 d x^2}}{35 c^3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]

[Out]

(2*b*d*x*Sqrt[d - c^2*d*x^2])/(35*c^3*Sqrt[1 - c^2*x^2]) + (b*d*x^3*Sqrt[d - c^2*d*x^2])/(105*c*Sqrt[1 - c^2*x
^2]) - (8*b*c*d*x^5*Sqrt[d - c^2*d*x^2])/(175*Sqrt[1 - c^2*x^2]) + (b*c^3*d*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqrt[
1 - c^2*x^2]) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d) + ((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c
*x]))/(7*c^4*d^2)

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])

Rule 4691

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], Int[x^m*(d + e*x^2)^p, x], x] - Dist[(b*c*d^(p - 1/2)*Sqrt[d +
e*x^2])/Sqrt[1 - c^2*x^2], Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x
] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int x^3 \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac{\left (b c d \sqrt{d-c^2 d x^2}\right ) \int \frac{\left (-2-5 c^2 x^2\right ) \left (1-c^2 x^2\right )^2}{35 c^4} \, dx}{\sqrt{1-c^2 x^2}}+\left (a+b \sin ^{-1}(c x)\right ) \int x^3 \left (d-c^2 d x^2\right )^{3/2} \, dx\\ &=-\frac{\left (b d \sqrt{d-c^2 d x^2}\right ) \int \left (-2-5 c^2 x^2\right ) \left (1-c^2 x^2\right )^2 \, dx}{35 c^3 \sqrt{1-c^2 x^2}}+\frac{1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname{Subst}\left (\int x \left (d-c^2 d x\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac{\left (b d \sqrt{d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+8 c^4 x^4-5 c^6 x^6\right ) \, dx}{35 c^3 \sqrt{1-c^2 x^2}}+\frac{1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname{Subst}\left (\int \left (\frac{\left (d-c^2 d x\right )^{3/2}}{c^2}-\frac{\left (d-c^2 d x\right )^{5/2}}{c^2 d}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b d x \sqrt{d-c^2 d x^2}}{35 c^3 \sqrt{1-c^2 x^2}}+\frac{b d x^3 \sqrt{d-c^2 d x^2}}{105 c \sqrt{1-c^2 x^2}}-\frac{8 b c d x^5 \sqrt{d-c^2 d x^2}}{175 \sqrt{1-c^2 x^2}}+\frac{b c^3 d x^7 \sqrt{d-c^2 d x^2}}{49 \sqrt{1-c^2 x^2}}-\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d}+\frac{\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.129352, size = 126, normalized size = 0.56 \[ \frac{d \sqrt{d-c^2 d x^2} \left (-105 a \left (5 c^2 x^2+2\right ) \left (1-c^2 x^2\right )^{5/2}+b c x \left (75 c^6 x^6-168 c^4 x^4+35 c^2 x^2+210\right )-105 b \left (5 c^2 x^2+2\right ) \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)\right )}{3675 c^4 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]

[Out]

(d*Sqrt[d - c^2*d*x^2]*(-105*a*(1 - c^2*x^2)^(5/2)*(2 + 5*c^2*x^2) + b*c*x*(210 + 35*c^2*x^2 - 168*c^4*x^4 + 7
5*c^6*x^6) - 105*b*(1 - c^2*x^2)^(5/2)*(2 + 5*c^2*x^2)*ArcSin[c*x]))/(3675*c^4*Sqrt[1 - c^2*x^2])

________________________________________________________________________________________

Maple [C]  time = 0.264, size = 931, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x)

[Out]

a*(-1/7*x^2*(-c^2*d*x^2+d)^(5/2)/c^2/d-2/35/d/c^4*(-c^2*d*x^2+d)^(5/2))+b*(-1/6272*(-d*(c^2*x^2-1))^(1/2)*(64*
c^8*x^8-144*c^6*x^6-64*I*(-c^2*x^2+1)^(1/2)*x^7*c^7+104*c^4*x^4+112*I*(-c^2*x^2+1)^(1/2)*x^5*c^5-25*c^2*x^2-56
*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+7*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(I+7*arcsin(c*x))*d/c^4/(c^2*x^2-1)+1/3200*(-d*(c^
2*x^2-1))^(1/2)*(16*c^6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*
c^3-5*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(I+5*arcsin(c*x))*d/c^4/(c^2*x^2-1)+1/384*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4-
5*c^2*x^2-4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(I+3*arcsin(c*x))*d/c^4/(c^2*x^2-1)-3/1
28*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)+I)*d/c^4/(c^2*x^2-1)-3/128*(-d*(c^
2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)-I)*d/c^4/(c^2*x^2-1)+1/384*(-d*(c^2*x^2-1))^
(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+1)*(-I+3*arcsin(c*x))*d/c
^4/(c^2*x^2-1)+1/3200*(-d*(c^2*x^2-1))^(1/2)*(16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+16*c^6*x^6-20*I*(-c^2*x^2+1)^(1/
2)*x^3*c^3-28*c^4*x^4+5*I*(-c^2*x^2+1)^(1/2)*x*c+13*c^2*x^2-1)*(-I+5*arcsin(c*x))*d/c^4/(c^2*x^2-1)-1/6272*(-d
*(c^2*x^2-1))^(1/2)*(64*I*(-c^2*x^2+1)^(1/2)*x^7*c^7+64*c^8*x^8-112*I*(-c^2*x^2+1)^(1/2)*x^5*c^5-144*c^6*x^6+5
6*I*(-c^2*x^2+1)^(1/2)*x^3*c^3+104*c^4*x^4-7*I*(-c^2*x^2+1)^(1/2)*x*c-25*c^2*x^2+1)*(-I+7*arcsin(c*x))*d/c^4/(
c^2*x^2-1))

________________________________________________________________________________________

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(x^3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.93341, size = 427, normalized size = 1.88 \begin{align*} -\frac{{\left (75 \, b c^{7} d x^{7} - 168 \, b c^{5} d x^{5} + 35 \, b c^{3} d x^{3} + 210 \, b c d x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} + 105 \,{\left (5 \, a c^{8} d x^{8} - 13 \, a c^{6} d x^{6} + 9 \, a c^{4} d x^{4} + a c^{2} d x^{2} - 2 \, a d +{\left (5 \, b c^{8} d x^{8} - 13 \, b c^{6} d x^{6} + 9 \, b c^{4} d x^{4} + b c^{2} d x^{2} - 2 \, b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{3675 \,{\left (c^{6} x^{2} - c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

-1/3675*((75*b*c^7*d*x^7 - 168*b*c^5*d*x^5 + 35*b*c^3*d*x^3 + 210*b*c*d*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2
+ 1) + 105*(5*a*c^8*d*x^8 - 13*a*c^6*d*x^6 + 9*a*c^4*d*x^4 + a*c^2*d*x^2 - 2*a*d + (5*b*c^8*d*x^8 - 13*b*c^6*d
*x^6 + 9*b*c^4*d*x^4 + b*c^2*d*x^2 - 2*b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^6*x^2 - c^4)

________________________________________________________________________________________

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(x**3*(-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((-c^2*d*x^2 + d)^(3/2)*(b*arcsin(c*x) + a)*x^3, x)