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

Optimal. Leaf size=144 \[ \frac {\left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f+g) \log (1-c x)}{2 c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f-g) \log (c x+1)}{2 c^2 d \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4777, 637, 4761, 12, 633, 31} \[ \frac {\left (c^2 f x+g\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f+g) \log (1-c x)}{2 c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} (c f-g) \log (c x+1)}{2 c^2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((g + c^2*f*x)*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) + (b*(c*f + g)*Sqrt[1 - c^2*x^2]*Log[1 - c*x])
/(2*c^2*d*Sqrt[d - c^2*d*x^2]) + (b*(c*f - g)*Sqrt[1 - c^2*x^2]*Log[1 + c*x])/(2*c^2*d*Sqrt[d - c^2*d*x^2])

Rule 12

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

Rule 31

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

Rule 633

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 637

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 4761

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.56, size = 135, normalized size = 0.94 \[ \frac {\sqrt {1-c^2 x^2} \left ((c f-g) \left (2 b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-\cot \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )\right )+(c f+g) \left (\tan \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )+2 b \log \left (\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )\right )\right )}{2 c^2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - c^2*x^2]*((c*f - g)*(-((a + b*ArcSin[c*x])*Cot[(Pi + 2*ArcSin[c*x])/4]) + 2*b*Log[Sin[(Pi + 2*ArcSin
[c*x])/4]]) + (c*f + g)*(2*b*Log[Cos[(Pi + 2*ArcSin[c*x])/4]] + (a + b*ArcSin[c*x])*Tan[(Pi + 2*ArcSin[c*x])/4
])))/(2*c^2*d*Sqrt[d - c^2*d*x^2])

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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (a g x + a f + {\left (b g x + b f\right )} \arcsin \left (c x\right )\right )}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [C]  time = 0.65, size = 443, normalized size = 3.08 \[ \frac {a g}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {a f x}{d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f \arcsin \left (c x \right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x f}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) f}{c \,d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) f}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*g/c^2/d/(-c^2*d*x^2+d)^(1/2)+a*f/d*x/(-c^2*d*x^2+d)^(1/2)+I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c/d^
2/(c^2*x^2-1)*f*arcsin(c*x)-b*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*x*f-b*(-d*(c^2*x^2-1))^(1/2)/
c^2/d^2/(c^2*x^2-1)*arcsin(c*x)*g-b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)/c
/d^2/(c^2*x^2-1)*f+b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)/c^2/d^2/(c^2*x^2
-1)*g-b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*f-b*(-c^2*x
^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^2/d^2/(c^2*x^2-1)*ln(I*c*x+(-c^2*x^2+1)^(1/2)-I)*g

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b f x \arcsin \left (c x\right )}{\sqrt {-c^{2} d x^{2} + d} d} + \frac {a f x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {b f \log \left (x^{2} - \frac {1}{c^{2}}\right )}{2 \, c d^{\frac {3}{2}}} + \frac {\frac {1}{2} \, {\left (\sqrt {c x + 1} \sqrt {-c x + 1} c^{3} d^{2} {\left (\frac {2 \, x}{c^{2} d^{2}} - \frac {\log \left (c x + 1\right )}{c^{3} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} + 2 \, \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} b g}{\sqrt {c x + 1} \sqrt {-c x + 1} c^{2} d^{\frac {3}{2}}} + \frac {a g}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))*(f + g*x)/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

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