3.1.54 \(\int \frac {(f+g x)^3 (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\) [54]

Optimal. Leaf size=410 \[ -\frac {b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) (a+b \text {ArcSin}(c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 (a+b \text {ArcSin}(c x))}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b (c f-g) (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b g (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 g \sqrt {1-c^2 x^2} \log (1+c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 (c f+g) \sqrt {1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

2/3*(c*f-g)*(c*f+g)*(c^2*f*x+g)*(a+b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*(c^2*f*x+g)*(g*x+f)^2*(a+b*
arcsin(c*x))/c^2/d^2/(-c^2*x^2+1)/(-c^2*d*x^2+d)^(1/2)-1/6*b*(g*x+f)*(2*c^2*f*g*x+c^2*f^2+g^2)/c^3/d^2/(-c^2*x
^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+1/3*b*(c*f-g)*(c*f+g)^2*ln(-c*x+1)*(-c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^
(1/2)-1/12*b*g*(c*f+g)^2*ln(-c*x+1)*(-c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/12*b*(c*f-g)^2*g*ln(c*x+
1)*(-c^2*x^2+1)^(1/2)/c^4/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*b*(c*f-g)^2*(c*f+g)*ln(c*x+1)*(-c^2*x^2+1)^(1/2)/c^4/d^
2/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 410, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4861, 737, 651, 4845, 833, 647, 31} \begin {gather*} \frac {(f+g x)^2 \left (c^2 f x+g\right ) (a+b \text {ArcSin}(c x))}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {2 (c f+g) (c f-g) \left (c^2 f x+g\right ) (a+b \text {ArcSin}(c x))}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b g \sqrt {1-c^2 x^2} (c f-g)^2 \log (c x+1)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f+g) (c f-g)^2 \log (c x+1)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f+g)^2 (c f-g) \log (1-c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b g \sqrt {1-c^2 x^2} (c f+g)^2 \log (1-c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b (f+g x) \left (c^2 f^2+2 c^2 f g x+g^2\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((f + g*x)^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-1/6*(b*(f + g*x)*(c^2*f^2 + g^2 + 2*c^2*f*g*x))/(c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (2*(c*f - g
)*(c*f + g)*(g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) + ((g + c^2*f*x)*(f + g*x)^2*(a
+ b*ArcSin[c*x]))/(3*c^2*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (b*(c*f - g)*(c*f + g)^2*Sqrt[1 - c^2*x^2]*
Log[1 - c*x])/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) - (b*g*(c*f + g)^2*Sqrt[1 - c^2*x^2]*Log[1 - c*x])/(12*c^4*d^2*S
qrt[d - c^2*d*x^2]) + (b*(c*f - g)^2*g*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(12*c^4*d^2*Sqrt[d - c^2*d*x^2]) + (b*(
c*f - g)^2*(c*f + g)*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(3*c^4*d^2*Sqrt[d - c^2*d*x^2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 4845

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

Rule 4861

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

Rubi steps

\begin {align*} \int \frac {(f+g x)^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \left (\frac {\left (g+c^2 f x\right ) (f+g x)^2}{3 c^2 \left (1-c^2 x^2\right )^2}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right )}{3 c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {\left (g+c^2 f x\right ) (f+g x)^2}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (2 b (c f-g) (c f+g) \sqrt {1-c^2 x^2}\right ) \int \frac {g+c^2 f x}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {g \left (c^2 f^2+g^2\right )+2 c^2 f g^2 x}{1-c^2 x^2} \, dx}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b (c f-g)^2 (c f+g) \sqrt {1-c^2 x^2}\right ) \int \frac {1}{-c-c^2 x} \, dx}{3 c^2 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b (c f-g) (c f+g)^2 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{c-c^2 x} \, dx}{3 c^2 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b (c f-g) (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 (c f+g) \sqrt {1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b (c f-g)^2 g \sqrt {1-c^2 x^2}\right ) \int \frac {1}{-c-c^2 x} \, dx}{12 c^2 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b g (c f+g)^2 \sqrt {1-c^2 x^2}\right ) \int \frac {1}{c-c^2 x} \, dx}{12 c^2 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b (f+g x) \left (c^2 f^2+g^2+2 c^2 f g x\right )}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 (c f-g) (c f+g) \left (g+c^2 f x\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (g+c^2 f x\right ) (f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b (c f-g) (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}-\frac {b g (c f+g)^2 \sqrt {1-c^2 x^2} \log (1-c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 g \sqrt {1-c^2 x^2} \log (1+c x)}{12 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g)^2 (c f+g) \sqrt {1-c^2 x^2} \log (1+c x)}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.81, size = 366, normalized size = 0.89 \begin {gather*} \frac {\sqrt {d-c^2 d x^2} \left (i b c g \left (3 c^2 f^2-5 g^2\right ) \left (1-c^2 x^2\right )^{3/2} F\left (\left .i \sinh ^{-1}\left (\sqrt {-c^2} x\right )\right |1\right )-\sqrt {-c^2} \left (-6 a c^2 f^2 g+4 a g^3-6 a c^4 f^3 x-6 a c^2 g^3 x^2+4 a c^6 f^3 x^3-6 a c^4 f g^2 x^3+b c^3 f^3 \sqrt {1-c^2 x^2}+3 b c f g^2 \sqrt {1-c^2 x^2}+3 b c^3 f^2 g x \sqrt {1-c^2 x^2}+b c g^3 x \sqrt {1-c^2 x^2}+2 b \left (2 g^3+2 c^6 f^3 x^3-3 c^2 g \left (f^2+g^2 x^2\right )-3 c^4 f x \left (f^2+g^2 x^2\right )\right ) \text {ArcSin}(c x)-b c f \left (2 c^2 f^2-3 g^2\right ) \left (1-c^2 x^2\right )^{3/2} \log \left (-1+c^2 x^2\right )\right )\right )}{6 c^4 \sqrt {-c^2} d^3 \left (-1+c^2 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(I*b*c*g*(3*c^2*f^2 - 5*g^2)*(1 - c^2*x^2)^(3/2)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1] -
Sqrt[-c^2]*(-6*a*c^2*f^2*g + 4*a*g^3 - 6*a*c^4*f^3*x - 6*a*c^2*g^3*x^2 + 4*a*c^6*f^3*x^3 - 6*a*c^4*f*g^2*x^3 +
b*c^3*f^3*Sqrt[1 - c^2*x^2] + 3*b*c*f*g^2*Sqrt[1 - c^2*x^2] + 3*b*c^3*f^2*g*x*Sqrt[1 - c^2*x^2] + b*c*g^3*x*S
qrt[1 - c^2*x^2] + 2*b*(2*g^3 + 2*c^6*f^3*x^3 - 3*c^2*g*(f^2 + g^2*x^2) - 3*c^4*f*x*(f^2 + g^2*x^2))*ArcSin[c*
x] - b*c*f*(2*c^2*f^2 - 3*g^2)*(1 - c^2*x^2)^(3/2)*Log[-1 + c^2*x^2])))/(6*c^4*Sqrt[-c^2]*d^3*(-1 + c^2*x^2)^2
)

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Maple [C] Result contains complex when optimal does not.
time = 0.80, size = 5114, normalized size = 12.47

method	result	size
default	\(\text {Expression too large to display}\)	\(5114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*f^3*(1/(c^4*d^(5/2)*x^2 - c^2*d^(5/2)) + 2*log(c*x + 1)/(c^2*d^(5/2)) + 2*log(c*x - 1)/(c^2*d^(5/2)))
+ 1/3*b*f^3*(2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d))*arcsin(c*x) + 1/3*a*f^3*(2*x/(sqrt
(-c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d)) + 1/3*a*g^3*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((
-c^2*d*x^2 + d)^(3/2)*c^4*d)) - a*f*g^2*(x/(sqrt(-c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d))
+ integrate((b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^4
- 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d) + a*f^2*g/((-c^2*d*x^2 + d)^(3/2)*c^2*d)

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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((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-(a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)
*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)

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Sympy [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((g*x+f)**3*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

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Giac [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((g*x+f)^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x + f)^3*(b*arcsin(c*x) + a)/(-c^2*d*x^2 + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int(((f + g*x)^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2), x)

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