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

Optimal. Leaf size=370 \[ \frac{3}{8} d f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d f x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3 d f \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}-\frac{d g \left (1-c^2 x^2\right )^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2}+\frac{b c^3 d f x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{5 b c d f x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{b c^3 d g x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}-\frac{2 b c d g x^3 \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{b d g x \sqrt{d-c^2 d x^2}}{5 c \sqrt{1-c^2 x^2}} \]

[Out]

(b*d*g*x*Sqrt[d - c^2*d*x^2])/(5*c*Sqrt[1 - c^2*x^2]) - (5*b*c*d*f*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x
^2]) - (2*b*c*d*g*x^3*Sqrt[d - c^2*d*x^2])/(15*Sqrt[1 - c^2*x^2]) + (b*c^3*d*f*x^4*Sqrt[d - c^2*d*x^2])/(16*Sq
rt[1 - c^2*x^2]) + (b*c^3*d*g*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) + (3*d*f*x*Sqrt[d - c^2*d*x^2]*(
a + b*ArcSin[c*x]))/8 + (d*f*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/4 - (d*g*(1 - c^2*x^2)^2
*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(5*c^2) + (3*d*f*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(16*b*c*
Sqrt[1 - c^2*x^2])

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Rubi [A]  time = 0.326304, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {4777, 4763, 4649, 4647, 4641, 30, 14, 4677, 194} \[ \frac{3}{8} d f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d f x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3 d f \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}-\frac{d g \left (1-c^2 x^2\right )^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2}+\frac{b c^3 d f x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{5 b c d f x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{b c^3 d g x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}-\frac{2 b c d g x^3 \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{b d g x \sqrt{d-c^2 d x^2}}{5 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*g*x*Sqrt[d - c^2*d*x^2])/(5*c*Sqrt[1 - c^2*x^2]) - (5*b*c*d*f*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x
^2]) - (2*b*c*d*g*x^3*Sqrt[d - c^2*d*x^2])/(15*Sqrt[1 - c^2*x^2]) + (b*c^3*d*f*x^4*Sqrt[d - c^2*d*x^2])/(16*Sq
rt[1 - c^2*x^2]) + (b*c^3*d*g*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) + (3*d*f*x*Sqrt[d - c^2*d*x^2]*(
a + b*ArcSin[c*x]))/8 + (d*f*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/4 - (d*g*(1 - c^2*x^2)^2
*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(5*c^2) + (3*d*f*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(16*b*c*
Sqrt[1 - c^2*x^2])

Rule 4777

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

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

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 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 (f+g x) \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \int (f+g x) \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (d \sqrt{d-c^2 d x^2}\right ) \int \left (f \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+g x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{\left (d f \sqrt{d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}+\frac{\left (d g \sqrt{d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt{1-c^2 x^2}}\\ &=\frac{1}{4} d f x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{d g \left (1-c^2 x^2\right )^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2}+\frac{\left (3 d f \sqrt{d-c^2 d x^2}\right ) \int \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4 \sqrt{1-c^2 x^2}}-\frac{\left (b c d f \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{\left (b d g \sqrt{d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^2 \, dx}{5 c \sqrt{1-c^2 x^2}}\\ &=\frac{3}{8} d f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d f x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{d g \left (1-c^2 x^2\right )^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2}+\frac{\left (3 d f \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 (b c d f \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 (3 b c d f \sqrt{d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt{1-c^2 x^2}}+\frac{\left (b d g \sqrt{d-c^2 d x^2}\right ) \int \left (1-2 c^2 x^2+c^4 x^4\right ) \, dx}{5 c \sqrt{1-c^2 x^2}}\\ &=\frac{b d g x \sqrt{d-c^2 d x^2}}{5 c \sqrt{1-c^2 x^2}}-\frac{5 b c d f x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{2 b c d g x^3 \sqrt{d-c^2 d x^2}}{15 \sqrt{1-c^2 x^2}}+\frac{b c^3 d f x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}+\frac{b c^3 d g x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{3}{8} d f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d f x \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{d g \left (1-c^2 x^2\right )^2 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2}+\frac{3 d f \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.28579, size = 216, normalized size = 0.58 \[ \frac{d \sqrt{d-c^2 d x^2} \left (225 a^2 c f-30 a b \sqrt{1-c^2 x^2} \left (5 c^2 f x \left (2 c^2 x^2-5\right )+8 g \left (c^2 x^2-1\right )^2\right )+30 b \sin ^{-1}(c x) \left (15 a c f+b \sqrt{1-c^2 x^2} \left (5 c^2 f x \left (5-2 c^2 x^2\right )-8 g \left (c^2 x^2-1\right )^2\right )\right )+b^2 c x \left (75 c^2 f x \left (c^2 x^2-5\right )+16 g \left (3 c^4 x^4-10 c^2 x^2+15\right )\right )+225 b^2 c f \sin ^{-1}(c x)^2\right )}{1200 b c^2 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d - c^2*d*x^2]*(225*a^2*c*f - 30*a*b*Sqrt[1 - c^2*x^2]*(8*g*(-1 + c^2*x^2)^2 + 5*c^2*f*x*(-5 + 2*c^2*x
^2)) + b^2*c*x*(75*c^2*f*x*(-5 + c^2*x^2) + 16*g*(15 - 10*c^2*x^2 + 3*c^4*x^4)) + 30*b*(15*a*c*f + b*Sqrt[1 -
c^2*x^2]*(5*c^2*f*x*(5 - 2*c^2*x^2) - 8*g*(-1 + c^2*x^2)^2))*ArcSin[c*x] + 225*b^2*c*f*ArcSin[c*x]^2))/(1200*b
*c^2*Sqrt[1 - c^2*x^2])

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Maple [B]  time = 0.42, size = 698, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a*g/c^2/d*(-c^2*d*x^2+d)^(5/2)+1/4*a*f*x*(-c^2*d*x^2+d)^(3/2)+3/8*a*f*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a*f*d^
2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/5*b*(-d*(c^2*x^2-1))^(1/2)*g*d*c^4/(c^2*x^2-1)*
arcsin(c*x)*x^6+3/5*b*(-d*(c^2*x^2-1))^(1/2)*g*d*c^2/(c^2*x^2-1)*arcsin(c*x)*x^4-3/5*b*(-d*(c^2*x^2-1))^(1/2)*
g*d/(c^2*x^2-1)*arcsin(c*x)*x^2-1/4*b*(-d*(c^2*x^2-1))^(1/2)*f*d*c^4/(c^2*x^2-1)*arcsin(c*x)*x^5+7/8*b*(-d*(c^
2*x^2-1))^(1/2)*f*d*c^2/(c^2*x^2-1)*arcsin(c*x)*x^3-17/128*b*(-d*(c^2*x^2-1))^(1/2)*f*d/c/(c^2*x^2-1)*(-c^2*x^
2+1)^(1/2)-5/8*b*(-d*(c^2*x^2-1))^(1/2)*f*d/(c^2*x^2-1)*arcsin(c*x)*x-1/25*b*(-d*(c^2*x^2-1))^(1/2)*g*d*c^3/(c
^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^5+2/15*b*(-d*(c^2*x^2-1))^(1/2)*g*d*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^3-1/5*b*
(-d*(c^2*x^2-1))^(1/2)*g*d/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-1/16*b*(-d*(c^2*x^2-1))^(1/2)*f*d*c^3/(c^2*x^2-1
)*(-c^2*x^2+1)^(1/2)*x^4+5/16*b*(-d*(c^2*x^2-1))^(1/2)*f*d*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-3/16*b*(-d*(c^
2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*f*d+1/5*b*(-d*(c^2*x^2-1))^(1/2)*g*d/c^2/(c^2*x
^2-1)*arcsin(c*x)

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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((g*x+f)*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),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 (-{\left (a c^{2} d g x^{3} + a c^{2} d f x^{2} - a d g x - a d f +{\left (b c^{2} d g x^{3} + b c^{2} d f x^{2} - b d g x - b d f\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*c^2*d*g*x^3 + a*c^2*d*f*x^2 - a*d*g*x - a*d*f + (b*c^2*d*g*x^3 + b*c^2*d*f*x^2 - b*d*g*x - b*d*f)
*arcsin(c*x))*sqrt(-c^2*d*x^2 + d), 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((g*x+f)*(-c**2*d*x**2+d)**(3/2)*(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{3}{2}}{\left (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-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)*(g*x + f)*(b*arcsin(c*x) + a), x)