3.1.51 \(\int \frac {(f+g x) (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [51]

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

[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 = 0.13, 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 = {4861, 651, 4845, 12, 647, 31} \begin {gather*} \frac {\left (c^2 f x+g\right ) (a+b \text {ArcSin}(c x))}{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}} \end {gather*}

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 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 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) \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 = 0.37, size = 135, normalized size = 0.94 \begin {gather*} \frac {\sqrt {1-c^2 x^2} \left ((c f-g) \left (-\left ((a+b \text {ArcSin}(c x)) \cot \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+2 b \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )\right )+(c f+g) \left (2 b \log \left (\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+(a+b \text {ArcSin}(c x)) \tan \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )\right )}{2 c^2 d \sqrt {d-c^2 d x^2}} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 0.49, size = 444, normalized size = 3.08

method	result	size
default	\(a \left (\frac {g}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {f x}{d \sqrt {-c^{2} d \,x^{2}+d}}\right )+\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f \arcsin \left (c x \right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x f}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) f}{c \,d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) f}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right ) g}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}\)	\(444\)

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,method=_RETURNVERBOSE)

[Out]

a*(g/c^2/d/(-c^2*d*x^2+d)^(1/2)+f*x/d/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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)*(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [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)*(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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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