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

Optimal. Leaf size=315 \[ \frac{5 x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac{5 x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac{5 (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c \left (1-c^2 x^2\right )^{5/2}}+\frac{1}{6} x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 b c^3 x^4 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}-\frac{25 b c x^2 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac{b \sqrt{1-c^2 x^2} (c d x+d)^{5/2} (f-c f x)^{5/2}}{36 c} \]

[Out]

(-25*b*c*x^2*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (5*b*c^3*x^4*(d + c*d*x)^(5/2)*(f
- c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (b*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*Sqrt[1 - c^2*x^2])/(36*c) +
(x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/6 + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a +
b*ArcSin[c*x]))/(16*(1 - c^2*x^2)^2) + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/(24*(1 -
c^2*x^2)) + (5*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x])^2)/(32*b*c*(1 - c^2*x^2)^(5/2))

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Rubi [A]  time = 0.265264, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {4673, 4649, 4647, 4641, 30, 14, 261} \[ \frac{5 x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac{5 x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac{5 (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c \left (1-c^2 x^2\right )^{5/2}}+\frac{1}{6} x (c d x+d)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 b c^3 x^4 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}-\frac{25 b c x^2 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac{b \sqrt{1-c^2 x^2} (c d x+d)^{5/2} (f-c f x)^{5/2}}{36 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*x^2*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (5*b*c^3*x^4*(d + c*d*x)^(5/2)*(f
- c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (b*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*Sqrt[1 - c^2*x^2])/(36*c) +
(x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/6 + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a +
b*ArcSin[c*x]))/(16*(1 - c^2*x^2)^2) + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/(24*(1 -
c^2*x^2)) + (5*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x])^2)/(32*b*c*(1 - c^2*x^2)^(5/2))

Rule 4673

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[((d + e*x)^q*(f + g*x)^q)/(1 - c^2*x^2)^q, Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{\left ((d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (1-c^2 x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{5/2}}\\ &=\frac{1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{6 \left (1-c^2 x^2\right )^{5/2}}-\frac{\left (b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \left (1-c^2 x^2\right )^2 \, dx}{6 \left (1-c^2 x^2\right )^{5/2}}\\ &=\frac{b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac{\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{8 \left (1-c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{24 \left (1-c^2 x^2\right )^{5/2}}\\ &=\frac{b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac{5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac{\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{16 \left (1-c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{24 \left (1-c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \, dx}{16 \left (1-c^2 x^2\right )^{5/2}}\\ &=-\frac{25 b c x^2 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac{5 b c^3 x^4 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac{b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt{1-c^2 x^2}}{36 c}+\frac{1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac{5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac{5 (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c \left (1-c^2 x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.51267, size = 303, normalized size = 0.96 \[ \frac{d^2 f^2 \left (\sqrt{c d x+d} \sqrt{f-c f x} \left (384 a c^5 x^5 \sqrt{1-c^2 x^2}-1248 a c^3 x^3 \sqrt{1-c^2 x^2}+1584 a c x \sqrt{1-c^2 x^2}+270 b \cos \left (2 \sin ^{-1}(c x)\right )+27 b \cos \left (4 \sin ^{-1}(c x)\right )+2 b \cos \left (6 \sin ^{-1}(c x)\right )\right )-720 a \sqrt{d} \sqrt{f} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )+360 b \sqrt{c d x+d} \sqrt{f-c f x} \sin ^{-1}(c x)^2+12 b \sqrt{c d x+d} \sqrt{f-c f x} \left (45 \sin \left (2 \sin ^{-1}(c x)\right )+9 \sin \left (4 \sin ^{-1}(c x)\right )+\sin \left (6 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)\right )}{2304 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*f^2*(360*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*a*Sqrt[d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan
[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(15
84*a*c*x*Sqrt[1 - c^2*x^2] - 1248*a*c^3*x^3*Sqrt[1 - c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 - c^2*x^2] + 270*b*Cos[2*
ArcSin[c*x]] + 27*b*Cos[4*ArcSin[c*x]] + 2*b*Cos[6*ArcSin[c*x]]) + 12*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin
[c*x]*(45*Sin[2*ArcSin[c*x]] + 9*Sin[4*ArcSin[c*x]] + Sin[6*ArcSin[c*x]])))/(2304*c*Sqrt[1 - c^2*x^2])

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Maple [F]  time = 0.226, size = 0, normalized size = 0. \begin{align*} \int \left ( cdx+d \right ) ^{{\frac{5}{2}}} \left ( -cfx+f \right ) ^{{\frac{5}{2}}} \left ( a+b\arcsin \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),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((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(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^{4} d^{2} f^{2} x^{4} - 2 \, a c^{2} d^{2} f^{2} x^{2} + a d^{2} f^{2} +{\left (b c^{4} d^{2} f^{2} x^{4} - 2 \, b c^{2} d^{2} f^{2} x^{2} + b d^{2} f^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{c d x + d} \sqrt{-c f x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*c^4*d^2*f^2*x^4 - 2*a*c^2*d^2*f^2*x^2 + a*d^2*f^2 + (b*c^4*d^2*f^2*x^4 - 2*b*c^2*d^2*f^2*x^2 + b*d
^2*f^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*f*x + f), 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*d*x+d)**(5/2)*(-c*f*x+f)**(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 d x + d\right )}^{\frac{5}{2}}{\left (-c f x + f\right )}^{\frac{5}{2}}{\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((c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

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