\(\int \frac {x}{a+b \sin (c+d x^2)} \, dx\) [37]

Optimal result

Integrand size = 16, antiderivative size = 48 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\frac {\arctan \left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d} \] Output:

arctan((b+a*tan(1/2*d*x^2+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\frac {\arctan \left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d} \] Input:

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

Output:

ArcTan[(b + a*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]]/(Sqrt[a^2 - b^2]*d)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3860, 3042, 3139, 1083, 217}

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.

\(\displaystyle \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{2} \int \frac {1}{a+b \sin \left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {1}{a+b \sin \left (d x^2+c\right )}dx^2\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\int \frac {1}{a x^4+a+2 b \tan \left (\frac {1}{2} \left (d x^2+c\right )\right )}d\tan \left (\frac {1}{2} \left (d x^2+c\right )\right )}{d}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {2 \int \frac {1}{-x^4-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} \left (d x^2+c\right )\right )\right )}{d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\arctan \left (\frac {2 a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}\)

Input:

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

Output:

ArcTan[(2*b + 2*a*Tan[(c + d*x^2)/2])/(2*Sqrt[a^2 - b^2])]/(Sqrt[a^2 - b^2 
]*d)
 

Defintions of rubi rules used

rule 217

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}\)	\(48\)
default	\(\frac {\arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}\)	\(48\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d}\)	\(138\)

Input:

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

Output:

1/d/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x^2+1/2*c)+2*b)/(a^2-b^2)^(1 
/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 208, normalized size of antiderivative = 4.33 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x^{2} + c\right ) \sin \left (d x^{2} + c\right ) + b \cos \left (d x^{2} + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2}}\right )}{4 \, {\left (a^{2} - b^{2}\right )} d}, -\frac {\arctan \left (-\frac {a \sin \left (d x^{2} + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x^{2} + c\right )}\right )}{2 \, \sqrt {a^{2} - b^{2}} d}\right ] \] Input:

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

Output:

[-1/4*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x^2 + c)^2 - 2*a*b*sin(d*x 
^2 + c) - a^2 - b^2 + 2*(a*cos(d*x^2 + c)*sin(d*x^2 + c) + b*cos(d*x^2 + c 
))*sqrt(-a^2 + b^2))/(b^2*cos(d*x^2 + c)^2 - 2*a*b*sin(d*x^2 + c) - a^2 - 
b^2))/((a^2 - b^2)*d), -1/2*arctan(-(a*sin(d*x^2 + c) + b)/(sqrt(a^2 - b^2 
)*cos(d*x^2 + c)))/(sqrt(a^2 - b^2)*d)]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (37) = 74\).

Time = 3.51 (sec) , antiderivative size = 165, normalized size of antiderivative = 3.44 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\begin {cases} \frac {\tilde {\infty } x^{2}}{\sin {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} \right )}}{2 b d} & \text {for}\: a = 0 \\\frac {x^{2}}{2 \left (a + b \sin {\left (c \right )}\right )} & \text {for}\: d = 0 \\\frac {1}{b d \tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} - b d} & \text {for}\: a = - b \\- \frac {1}{b d \tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + b d} & \text {for}\: a = b \\\frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + \frac {b}{a} - \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{2 d \sqrt {- a^{2} + b^{2}}} - \frac {\log {\left (\tan {\left (\frac {c}{2} + \frac {d x^{2}}{2} \right )} + \frac {b}{a} + \frac {\sqrt {- a^{2} + b^{2}}}{a} \right )}}{2 d \sqrt {- a^{2} + b^{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*x**2/sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(tan(c/2 
+ d*x**2/2))/(2*b*d), Eq(a, 0)), (x**2/(2*(a + b*sin(c))), Eq(d, 0)), (1/( 
b*d*tan(c/2 + d*x**2/2) - b*d), Eq(a, -b)), (-1/(b*d*tan(c/2 + d*x**2/2) + 
 b*d), Eq(a, b)), (log(tan(c/2 + d*x**2/2) + b/a - sqrt(-a**2 + b**2)/a)/( 
2*d*sqrt(-a**2 + b**2)) - log(tan(c/2 + d*x**2/2) + b/a + sqrt(-a**2 + b** 
2)/a)/(2*d*sqrt(-a**2 + b**2)), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8078 vs. \(2 (43) = 86\).

Time = 23.68 (sec) , antiderivative size = 8078, normalized size of antiderivative = 168.29 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*arctan2(-2*(4*(a^2*b^4 - b^6)*cos(d*x^2 + 2*c)^4*cos(c)*sin(c) - 4*(a^ 
2*b^4 - b^6)*cos(c)*sin(d*x^2 + 2*c)^4*sin(c) - 4*((a^3*b^3 - a*b^5)*cos(c 
)^3 + 3*(a^3*b^3 - a*b^5)*cos(c)*sin(c)^2)*cos(d*x^2 + 2*c)^3 - 4*(3*(a^3* 
b^3 - a*b^5)*cos(c)^2*sin(c) + (a^3*b^3 - a*b^5)*sin(c)^3 + ((a^2*b^4 - b^ 
6)*cos(c)^2 - (a^2*b^4 - b^6)*sin(c)^2)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c) 
^3 + 4*((4*a^4*b^2 - 5*a^2*b^4 + b^6)*cos(c)^3*sin(c) + (4*a^4*b^2 - 5*a^2 
*b^4 + b^6)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c)^2 - 4*((4*a^4*b^2 - 5*a^2*b^ 
4 + b^6)*cos(c)^3*sin(c) + (4*a^4*b^2 - 5*a^2*b^4 + b^6)*cos(c)*sin(c)^3 + 
 3*((a^3*b^3 - a*b^5)*cos(c)^3 - (a^3*b^3 - a*b^5)*cos(c)*sin(c)^2)*cos(d* 
x^2 + 2*c))*sin(d*x^2 + 2*c)^2 - 4*((2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)^5 
 + 2*(2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)^3*sin(c)^2 + (2*a^5*b - 3*a^3*b^ 
3 + a*b^5)*cos(c)*sin(c)^4)*cos(d*x^2 + 2*c) - 4*((2*a^5*b - 3*a^3*b^3 + a 
*b^5)*cos(c)^4*sin(c) + 2*(2*a^5*b - 3*a^3*b^3 + a*b^5)*cos(c)^2*sin(c)^3 
+ (2*a^5*b - 3*a^3*b^3 + a*b^5)*sin(c)^5 + ((a^2*b^4 - b^6)*cos(c)^2 - (a^ 
2*b^4 - b^6)*sin(c)^2)*cos(d*x^2 + 2*c)^3 - 3*((a^3*b^3 - a*b^5)*cos(c)^2* 
sin(c) - (a^3*b^3 - a*b^5)*sin(c)^3)*cos(d*x^2 + 2*c)^2 + ((4*a^4*b^2 - 5* 
a^2*b^4 + b^6)*cos(c)^4 - (4*a^4*b^2 - 5*a^2*b^4 + b^6)*sin(c)^4)*cos(d*x^ 
2 + 2*c))*sin(d*x^2 + 2*c) + (b^5*cos(d*x^2 + 2*c)^5*cos(c) - 4*a*b^4*cos( 
d*x^2 + 2*c)^4*cos(c)*sin(c) + b^5*sin(d*x^2 + 2*c)^5*sin(c) + (b^5*cos(d* 
x^2 + 2*c)*cos(c) + 4*a*b^4*cos(c)*sin(c))*sin(d*x^2 + 2*c)^4 + 2*((2*a...
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.31 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\frac {\pi \left \lfloor \frac {d x^{2} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} d} \] Input:

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

Output:

(pi*floor(1/2*(d*x^2 + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x^2 + 1/2 
*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*d)
 

Mupad [B] (verification not implemented)

Time = 40.71 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.67 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=-\frac {\ln \left (-x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}-\frac {2\,x\,\left (b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {b-a}}\right )-\ln \left (-x\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}+\frac {2\,x\,\left (b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{d\,x^2\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {b-a}}\right )}{2\,d\,\sqrt {a+b}\,\sqrt {b-a}} \] Input:

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

Output:

-(log(- x*exp(d*x^2*1i)*exp(c*1i)*2i - (2*x*(b*1i + a*exp(d*x^2*1i)*exp(c* 
1i)))/((a + b)^(1/2)*(b - a)^(1/2))) - log((2*x*(b*1i + a*exp(d*x^2*1i)*ex 
p(c*1i)))/((a + b)^(1/2)*(b - a)^(1/2)) - x*exp(d*x^2*1i)*exp(c*1i)*2i))/( 
2*d*(a + b)^(1/2)*(b - a)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {x}{a+b \sin \left (c+d x^2\right )} \, dx=\frac {\sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right )}{d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(sqrt(a**2 - b**2)*atan((tan((c + d*x**2)/2)*a + b)/sqrt(a**2 - b**2)))/(d 
*(a**2 - b**2))