3.1.57 \(\int \frac {a+b \arcsin (c x)}{(f+g x) (d-c^2 d x^2)^{5/2}} \, dx\) [57]

3.1.57.1 Optimal result

Integrand size = 31, antiderivative size = 1300 \[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {(c f-2 g) \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{4 d^2 (c f-g)^2 \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{12 d^2 (c f-g) \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{24 d^2 (c f-g) \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{24 d^2 (c f-g) \sqrt {d-c^2 d x^2}}-\frac {i g^4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d^2 (c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {i g^4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{d^2 (c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )\right )}{6 d^2 (c f+g) \sqrt {d-c^2 d x^2}}+\frac {b (c f+2 g) \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )\right )}{2 d^2 (c f+g)^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-2 g) \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )\right )}{2 d^2 (c f-g)^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )\right )}{6 d^2 (c f-g) \sqrt {d-c^2 d x^2}}-\frac {b g^4 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d^2 (c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {b g^4 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{d^2 (c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{24 d^2 (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{12 d^2 (c f+g) \sqrt {d-c^2 d x^2}}+\frac {(c f+2 g) \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{4 d^2 (c f+g)^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{24 d^2 (c f+g) \sqrt {d-c^2 d x^2}} \]

output

-1/4*(c*f-2*g)*(a+b*arcsin(c*x))*cot(1/4*Pi+1/2*arcsin(c*x))*(-c^2*x^2+1)^ 
(1/2)/d^2/(c*f-g)^2/(-c^2*d*x^2+d)^(1/2)-1/12*(a+b*arcsin(c*x))*cot(1/4*Pi 
+1/2*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/d^2/(c*f-g)/(-c^2*d*x^2+d)^(1/2)-1/24 
*b*csc(1/4*Pi+1/2*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/d^2/(c*f-g)/(-c^2*d*x^ 
2+d)^(1/2)-1/24*(a+b*arcsin(c*x))*cot(1/4*Pi+1/2*arcsin(c*x))*csc(1/4*Pi+1 
/2*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/d^2/(c*f-g)/(-c^2*d*x^2+d)^(1/2)+1/6* 
b*ln(cos(1/4*Pi+1/2*arcsin(c*x)))*(-c^2*x^2+1)^(1/2)/d^2/(c*f+g)/(-c^2*d*x 
^2+d)^(1/2)+1/2*b*(c*f+2*g)*ln(cos(1/4*Pi+1/2*arcsin(c*x)))*(-c^2*x^2+1)^( 
1/2)/d^2/(c*f+g)^2/(-c^2*d*x^2+d)^(1/2)+1/2*b*(c*f-2*g)*ln(sin(1/4*Pi+1/2* 
arcsin(c*x)))*(-c^2*x^2+1)^(1/2)/d^2/(c*f-g)^2/(-c^2*d*x^2+d)^(1/2)+1/6*b* 
ln(sin(1/4*Pi+1/2*arcsin(c*x)))*(-c^2*x^2+1)^(1/2)/d^2/(c*f-g)/(-c^2*d*x^2 
+d)^(1/2)-I*g^4*(a+b*arcsin(c*x))*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f 
-(c^2*f^2-g^2)^(1/2)))*(-c^2*x^2+1)^(1/2)/d^2/(c^2*f^2-g^2)^(5/2)/(-c^2*d* 
x^2+d)^(1/2)+I*g^4*(a+b*arcsin(c*x))*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/( 
c*f+(c^2*f^2-g^2)^(1/2)))*(-c^2*x^2+1)^(1/2)/d^2/(c^2*f^2-g^2)^(5/2)/(-c^2 
*d*x^2+d)^(1/2)-b*g^4*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f 
^2-g^2)^(1/2)))*(-c^2*x^2+1)^(1/2)/d^2/(c^2*f^2-g^2)^(5/2)/(-c^2*d*x^2+d)^ 
(1/2)+b*g^4*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2-g^2)^(1 
/2)))*(-c^2*x^2+1)^(1/2)/d^2/(c^2*f^2-g^2)^(5/2)/(-c^2*d*x^2+d)^(1/2)-1/24 
*b*sec(1/4*Pi+1/2*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/d^2/(c*f+g)/(-c^2*d...
 

3.1.57.2 Mathematica [A] (warning: unable to verify)

Time = 12.89 (sec) , antiderivative size = 2078, normalized size of antiderivative = 1.60 \[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Result too large to show} \]

input

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

output

Sqrt[-(d*(-1 + c^2*x^2))]*((a*g - a*c^2*f*x)/(3*d^3*(-(c^2*f^2) + g^2)*(-1 
 + c^2*x^2)^2) + (-3*a*g^3 - 2*a*c^4*f^3*x + 5*a*c^2*f*g^2*x)/(3*d^3*(-(c^ 
2*f^2) + g^2)^2*(-1 + c^2*x^2))) + (a*g^4*Log[f + g*x])/(d^(5/2)*(-(c*f) + 
 g)^2*(c*f + g)^2*Sqrt[-(c^2*f^2) + g^2]) - (a*g^4*Log[d*g + c^2*d*f*x + S 
qrt[d]*Sqrt[-(c^2*f^2) + g^2]*Sqrt[-(d*(-1 + c^2*x^2))]])/(d^(5/2)*(-(c*f) 
 + g)^2*(c*f + g)^2*Sqrt[-(c^2*f^2) + g^2]) + (b*((g*(-(c^2*f^2) + 7*g^2)* 
(1 - c^2*x^2)^(3/2)*ArcSin[c*x])/(6*(-(c^2*f^2) + g^2)^2*(d*(1 - c^2*x^2)) 
^(3/2)) + ((4*c*f + 7*g)*(1 - c^2*x^2)^(3/2)*Log[Cos[ArcSin[c*x]/2] - Sin[ 
ArcSin[c*x]/2]])/(6*(c*f + g)^2*(d*(1 - c^2*x^2))^(3/2)) + ((4*c*f - 7*g)* 
(1 - c^2*x^2)^(3/2)*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])/(6*(c*f 
- g)^2*(d*(1 - c^2*x^2))^(3/2)) + (g^4*(1 - c^2*x^2)^(3/2)*((Pi*ArcTan[(g 
+ c*f*Tan[ArcSin[c*x]/2])/Sqrt[c^2*f^2 - g^2]])/Sqrt[c^2*f^2 - g^2] + (2*( 
Pi/2 - ArcSin[c*x])*ArcTanh[((c*f + g)*Cot[(Pi/2 - ArcSin[c*x])/2])/Sqrt[- 
(c^2*f^2) + g^2]] - 2*ArcCos[-((c*f)/g)]*ArcTanh[((-(c*f) + g)*Tan[(Pi/2 - 
 ArcSin[c*x])/2])/Sqrt[-(c^2*f^2) + g^2]] + (ArcCos[-((c*f)/g)] - (2*I)*(A 
rcTanh[((c*f + g)*Cot[(Pi/2 - ArcSin[c*x])/2])/Sqrt[-(c^2*f^2) + g^2]] - A 
rcTanh[((-(c*f) + g)*Tan[(Pi/2 - ArcSin[c*x])/2])/Sqrt[-(c^2*f^2) + g^2]]) 
)*Log[Sqrt[-(c^2*f^2) + g^2]/(Sqrt[2]*E^((I/2)*(Pi/2 - ArcSin[c*x]))*Sqrt[ 
g]*Sqrt[c*f + c*g*x])] + (ArcCos[-((c*f)/g)] + (2*I)*(ArcTanh[((c*f + g)*C 
ot[(Pi/2 - ArcSin[c*x])/2])/Sqrt[-(c^2*f^2) + g^2]] - ArcTanh[((-(c*f) ...
 

3.1.57.3 Rubi [A] (verified)

Time = 2.08 (sec) , antiderivative size = 821, normalized size of antiderivative = 0.63, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5276, 5274, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2} (f+g x)} \, dx\)

\(\Big \downarrow \) 5276

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {a+b \arcsin (c x)}{(f+g x) \left (1-c^2 x^2\right )^{5/2}}dx}{d^2 \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 5274

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \left (\frac {(a+b \arcsin (c x)) g^4}{(g-c f)^2 (c f+g)^2 (f+g x) \sqrt {1-c^2 x^2}}-\frac {c (c f+2 g) (a+b \arcsin (c x))}{4 (c f+g)^2 (c x-1) \sqrt {1-c^2 x^2}}+\frac {c (c f-2 g) (a+b \arcsin (c x))}{4 (c f-g)^2 (c x+1) \sqrt {1-c^2 x^2}}+\frac {c (a+b \arcsin (c x))}{4 (c f+g) (c x-1)^2 \sqrt {1-c^2 x^2}}+\frac {c (a+b \arcsin (c x))}{4 (c f-g) (c x+1)^2 \sqrt {1-c^2 x^2}}\right )dx}{d^2 \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {i (a+b \arcsin (c x)) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) g^4}{(c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2}}+\frac {i (a+b \arcsin (c x)) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) g^4}{(c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2}}-\frac {b \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) g^4}{(c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2}}+\frac {b \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) g^4}{(c f-g)^2 (c f+g)^2 \sqrt {c^2 f^2-g^2}}-\frac {(a+b \arcsin (c x)) \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{24 (c f-g)}-\frac {b \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{24 (c f-g)}-\frac {b \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{24 (c f+g)}-\frac {(a+b \arcsin (c x)) \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{12 (c f-g)}-\frac {(c f-2 g) (a+b \arcsin (c x)) \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{4 (c f-g)^2}+\frac {b (c f+2 g) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )\right )}{2 (c f+g)^2}+\frac {b \log \left (\cos \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )\right )}{6 (c f+g)}+\frac {b \log \left (\sin \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )\right )}{6 (c f-g)}+\frac {b (c f-2 g) \log \left (\sin \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )\right )}{2 (c f-g)^2}+\frac {(a+b \arcsin (c x)) \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{24 (c f+g)}+\frac {(c f+2 g) (a+b \arcsin (c x)) \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{4 (c f+g)^2}+\frac {(a+b \arcsin (c x)) \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{12 (c f+g)}\right )}{d^2 \sqrt {d-c^2 d x^2}}\)

input

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

output

(Sqrt[1 - c^2*x^2]*(-1/4*((c*f - 2*g)*(a + b*ArcSin[c*x])*Cot[Pi/4 + ArcSi 
n[c*x]/2])/(c*f - g)^2 - ((a + b*ArcSin[c*x])*Cot[Pi/4 + ArcSin[c*x]/2])/( 
12*(c*f - g)) - (b*Csc[Pi/4 + ArcSin[c*x]/2]^2)/(24*(c*f - g)) - ((a + b*A 
rcSin[c*x])*Cot[Pi/4 + ArcSin[c*x]/2]*Csc[Pi/4 + ArcSin[c*x]/2]^2)/(24*(c* 
f - g)) - (I*g^4*(a + b*ArcSin[c*x])*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f 
- Sqrt[c^2*f^2 - g^2])])/((c*f - g)^2*(c*f + g)^2*Sqrt[c^2*f^2 - g^2]) + ( 
I*g^4*(a + b*ArcSin[c*x])*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2* 
f^2 - g^2])])/((c*f - g)^2*(c*f + g)^2*Sqrt[c^2*f^2 - g^2]) + (b*Log[Cos[P 
i/4 + ArcSin[c*x]/2]])/(6*(c*f + g)) + (b*(c*f + 2*g)*Log[Cos[Pi/4 + ArcSi 
n[c*x]/2]])/(2*(c*f + g)^2) + (b*(c*f - 2*g)*Log[Sin[Pi/4 + ArcSin[c*x]/2] 
])/(2*(c*f - g)^2) + (b*Log[Sin[Pi/4 + ArcSin[c*x]/2]])/(6*(c*f - g)) - (b 
*g^4*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/((c* 
f - g)^2*(c*f + g)^2*Sqrt[c^2*f^2 - g^2]) + (b*g^4*PolyLog[2, (I*E^(I*ArcS 
in[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/((c*f - g)^2*(c*f + g)^2*Sqrt[c^ 
2*f^2 - g^2]) - (b*Sec[Pi/4 + ArcSin[c*x]/2]^2)/(24*(c*f + g)) + ((a + b*A 
rcSin[c*x])*Tan[Pi/4 + ArcSin[c*x]/2])/(12*(c*f + g)) + ((c*f + 2*g)*(a + 
b*ArcSin[c*x])*Tan[Pi/4 + ArcSin[c*x]/2])/(4*(c*f + g)^2) + ((a + b*ArcSin 
[c*x])*Sec[Pi/4 + ArcSin[c*x]/2]^2*Tan[Pi/4 + ArcSin[c*x]/2])/(24*(c*f + g 
))))/(d^2*Sqrt[d - c^2*d*x^2])
 

3.1.57.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5274

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

rule 5276

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

3.1.57.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7970 vs. \(2 (1102 ) = 2204\).

Time = 1.50 (sec) , antiderivative size = 7971, normalized size of antiderivative = 6.13

method	result	size
default	\(\text {Expression too large to display}\)	\(7971\)
parts	\(\text {Expression too large to display}\)	\(7971\)

input

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

output

result too large to display
 

3.1.57.5 Fricas [F]

\[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}} \,d x } \]

input

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

output

integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*g*x^7 + c^6*d^ 
3*f*x^6 - 3*c^4*d^3*g*x^5 - 3*c^4*d^3*f*x^4 + 3*c^2*d^3*g*x^3 + 3*c^2*d^3* 
f*x^2 - d^3*g*x - d^3*f), x)
 

3.1.57.6 Sympy [F]

\[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}} \left (f + g x\right )}\, dx \]

input

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

output

Integral((a + b*asin(c*x))/((-d*(c*x - 1)*(c*x + 1))**(5/2)*(f + g*x)), x)
 

3.1.57.7 Maxima [F]

\[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}} \,d x } \]

input

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

output

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

3.1.57.8 Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

3.1.57.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin (c x)}{(f+g x) \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\left (f+g\,x\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

input

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

output

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