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

Optimal. Leaf size=202 \[ -\frac{\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^2 d}-\frac{b c^5 d^2 x^7 \sqrt{d-c^2 d x^2}}{49 \sqrt{1-c^2 x^2}}+\frac{3 b c^3 d^2 x^5 \sqrt{d-c^2 d x^2}}{35 \sqrt{1-c^2 x^2}}-\frac{b c d^2 x^3 \sqrt{d-c^2 d x^2}}{7 \sqrt{1-c^2 x^2}}+\frac{b d^2 x \sqrt{d-c^2 d x^2}}{7 c \sqrt{1-c^2 x^2}} \]

[Out]

(b*d^2*x*Sqrt[d - c^2*d*x^2])/(7*c*Sqrt[1 - c^2*x^2]) - (b*c*d^2*x^3*Sqrt[d - c^2*d*x^2])/(7*Sqrt[1 - c^2*x^2]
) + (3*b*c^3*d^2*x^5*Sqrt[d - c^2*d*x^2])/(35*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqr
t[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(7*c^2*d)

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Rubi [A]  time = 0.0907841, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {4677, 194} \[ -\frac{\left (d-c^2 d x^2\right )^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^2 d}-\frac{b c^5 d^2 x^7 \sqrt{d-c^2 d x^2}}{49 \sqrt{1-c^2 x^2}}+\frac{3 b c^3 d^2 x^5 \sqrt{d-c^2 d x^2}}{35 \sqrt{1-c^2 x^2}}-\frac{b c d^2 x^3 \sqrt{d-c^2 d x^2}}{7 \sqrt{1-c^2 x^2}}+\frac{b d^2 x \sqrt{d-c^2 d x^2}}{7 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*x*Sqrt[d - c^2*d*x^2])/(7*c*Sqrt[1 - c^2*x^2]) - (b*c*d^2*x^3*Sqrt[d - c^2*d*x^2])/(7*Sqrt[1 - c^2*x^2]
) + (3*b*c^3*d^2*x^5*Sqrt[d - c^2*d*x^2])/(35*Sqrt[1 - c^2*x^2]) - (b*c^5*d^2*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqr
t[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(7*c^2*d)

Rule 4677

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

Rule 194

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

Rubi steps

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

Mathematica [A]  time = 0.0694342, size = 93, normalized size = 0.46 \[ \frac{d^2 \sqrt{d-c^2 d x^2} \left (\left (c^2 x^2-1\right )^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{b c \left (-\frac{1}{7} c^6 x^7+\frac{3 c^4 x^5}{5}-c^2 x^3+x\right )}{\sqrt{1-c^2 x^2}}\right )}{7 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.223, size = 921, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a/c^2/d*(-c^2*d*x^2+d)^(7/2)+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^2/c^2/(c^2*x^2-1)-1/640*(-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^2/c^2/(c^2*x^2-1)+1/128*(-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^2/c^2/(c^2*x^2-1)-5/128*(-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^2/c^2/(c^2*x^2-1)-5/128*(-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^2/c^2/(c^2*x^2-1)+1/128*(-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^2/c^2/(c^2*x^2-1)-1/640*(-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^2/c^2/(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+56*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^2/c^2/(c^2*x^2-1))

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Maxima [A]  time = 1.6537, size = 132, normalized size = 0.65 \begin{align*} -\frac{{\left (-c^{2} d x^{2} + d\right )}^{\frac{7}{2}} b \arcsin \left (c x\right )}{7 \, c^{2} d} - \frac{{\left (-c^{2} d x^{2} + d\right )}^{\frac{7}{2}} a}{7 \, c^{2} d} - \frac{{\left (5 \, c^{6} d^{\frac{7}{2}} x^{7} - 21 \, c^{4} d^{\frac{7}{2}} x^{5} + 35 \, c^{2} d^{\frac{7}{2}} x^{3} - 35 \, d^{\frac{7}{2}} x\right )} b}{245 \, c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(-c^2*d*x^2 + d)^(7/2)*b*arcsin(c*x)/(c^2*d) - 1/7*(-c^2*d*x^2 + d)^(7/2)*a/(c^2*d) - 1/245*(5*c^6*d^(7/2
)*x^7 - 21*c^4*d^(7/2)*x^5 + 35*c^2*d^(7/2)*x^3 - 35*d^(7/2)*x)*b/(c*d)

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Fricas [A]  time = 2.25337, size = 448, normalized size = 2.22 \begin{align*} \frac{{\left (5 \, b c^{7} d^{2} x^{7} - 21 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3} - 35 \, b c d^{2} x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} + 35 \,{\left (a c^{8} d^{2} x^{8} - 4 \, a c^{6} d^{2} x^{6} + 6 \, a c^{4} d^{2} x^{4} - 4 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{8} d^{2} x^{8} - 4 \, b c^{6} d^{2} x^{6} + 6 \, b c^{4} d^{2} x^{4} - 4 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{245 \,{\left (c^{4} x^{2} - c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/245*((5*b*c^7*d^2*x^7 - 21*b*c^5*d^2*x^5 + 35*b*c^3*d^2*x^3 - 35*b*c*d^2*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x
^2 + 1) + 35*(a*c^8*d^2*x^8 - 4*a*c^6*d^2*x^6 + 6*a*c^4*d^2*x^4 - 4*a*c^2*d^2*x^2 + a*d^2 + (b*c^8*d^2*x^8 - 4
*b*c^6*d^2*x^6 + 6*b*c^4*d^2*x^4 - 4*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^4*x^2 - c^2)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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