3.2.83 \(\int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx\) [183]

3.2.83.1 Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=-\frac {a+b \arcsin (c+d x)}{d e^2 (c+d x)}-\frac {b \text {arctanh}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2} \]

output

(-a-b*arcsin(d*x+c))/d/e^2/(d*x+c)-b*arctanh((1-(d*x+c)^2)^(1/2))/d/e^2
 

3.2.83.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=-\frac {\frac {a+b \arcsin (c+d x)}{c+d x}+b \text {arctanh}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2} \]

input

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

output

-(((a + b*ArcSin[c + d*x])/(c + d*x) + b*ArcTanh[Sqrt[1 - (c + d*x)^2]])/( 
d*e^2))
 

3.2.83.3 Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5304, 27, 5138, 243, 73, 219}

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.

\(\displaystyle \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx\)

\(\Big \downarrow \) 5304

\(\displaystyle \frac {\int \frac {a+b \arcsin (c+d x)}{e^2 (c+d x)^2}d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a+b \arcsin (c+d x)}{(c+d x)^2}d(c+d x)}{d e^2}\)

\(\Big \downarrow \) 5138

\(\displaystyle \frac {b \int \frac {1}{(c+d x) \sqrt {1-(c+d x)^2}}d(c+d x)-\frac {a+b \arcsin (c+d x)}{c+d x}}{d e^2}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{2} b \int \frac {1}{\sqrt {-c-d x+1} (c+d x)^2}d(c+d x)^2-\frac {a+b \arcsin (c+d x)}{c+d x}}{d e^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-b \int \frac {1}{1-(c+d x)^4}d\sqrt {-c-d x+1}-\frac {a+b \arcsin (c+d x)}{c+d x}}{d e^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {a+b \arcsin (c+d x)}{c+d x}-b \text {arctanh}\left (\sqrt {-c-d x+1}\right )}{d e^2}\)

input

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

output

(-((a + b*ArcSin[c + d*x])/(c + d*x)) - b*ArcTanh[Sqrt[1 - c - d*x]])/(d*e 
^2)
 

3.2.83.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 5138

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n 
/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 
*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
 

rule 5304

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

3.2.83.4 Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d}\)	\(56\)
default	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d}\)	\(56\)
parts	\(-\frac {a}{e^{2} \left (d x +c \right ) d}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2} d}\)	\(58\)

input

int((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x,method=_RETURNVERBOSE)
 

output

1/d*(-a/e^2/(d*x+c)+b/e^2*(-1/(d*x+c)*arcsin(d*x+c)-arctanh(1/(1-(d*x+c)^2 
)^(1/2))))
 

3.2.83.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.98 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=-\frac {2 \, b \arcsin \left (d x + c\right ) + {\left (b d x + b c\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} + 1\right ) - {\left (b d x + b c\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} - 1\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]

input

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

output

-1/2*(2*b*arcsin(d*x + c) + (b*d*x + b*c)*log(sqrt(-d^2*x^2 - 2*c*d*x - c^ 
2 + 1) + 1) - (b*d*x + b*c)*log(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1) - 1) + 
2*a)/(d^2*e^2*x + c*d*e^2)
 

3.2.83.6 Sympy [F]

\[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=\frac {\int \frac {a}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b \operatorname {asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]

input

integrate((a+b*asin(d*x+c))/(d*e*x+c*e)**2,x)
 

output

(Integral(a/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b*asin(c + d*x)/(c 
**2 + 2*c*d*x + d**2*x**2), x))/e**2
 

3.2.83.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=\text {Exception raised: ValueError} \]

input

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

3.2.83.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (49) = 98\).

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.12 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^2} \, dx=-\frac {1}{2} \, b d e^{2} {\left (\frac {\log \left (\sqrt {-\frac {{\left (d e x + c e\right )}^{2}}{e^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {{\left (d e x + c e\right )}^{2}}{e^{2}} + 1} + 1\right )}{d^{2} e^{4}} + \frac {2 \, \arcsin \left (d x + c\right )}{{\left (d e x + c e\right )} d^{2} e^{3}}\right )} - \frac {a}{{\left (d e x + c e\right )} d e} \]

input

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x, algorithm="giac")
 

output

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

3.2.83.9 Mupad [F(-1)]

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

input

int((a + b*asin(c + d*x))/(c*e + d*e*x)^2,x)
 

output

int((a + b*asin(c + d*x))/(c*e + d*e*x)^2, x)