Integrand size = 18, antiderivative size = 74 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=-\frac {1}{4} b^2 \cos (2 c) \operatorname {CosIntegral}\left (2 d x^2\right )+\frac {1}{2} \left (2 a^2+b^2\right ) \log (x)+a b \operatorname {CosIntegral}\left (d x^2\right ) \sin (c)+a b \cos (c) \text {Si}\left (d x^2\right )+\frac {1}{4} b^2 \sin (2 c) \text {Si}\left (2 d x^2\right ) \] Output:
-1/4*b^2*cos(2*c)*Ci(2*d*x^2)+1/2*(2*a^2+b^2)*ln(x)+a*b*Ci(d*x^2)*sin(c)+a *b*cos(c)*Si(d*x^2)+1/4*b^2*sin(2*c)*Si(2*d*x^2)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=\frac {1}{2} \left (2 a^2+b^2\right ) \log (x)-\frac {1}{4} b \left (b \cos (2 c) \operatorname {CosIntegral}\left (2 d x^2\right )-4 a \operatorname {CosIntegral}\left (d x^2\right ) \sin (c)-4 a \cos (c) \text {Si}\left (d x^2\right )-b \sin (2 c) \text {Si}\left (2 d x^2\right )\right ) \] Input:
Integrate[(a + b*Sin[c + d*x^2])^2/x,x]
Output:
((2*a^2 + b^2)*Log[x])/2 - (b*(b*Cos[2*c]*CosIntegral[2*d*x^2] - 4*a*CosIn tegral[d*x^2]*Sin[c] - 4*a*Cos[c]*SinIntegral[d*x^2] - b*Sin[2*c]*SinInteg ral[2*d*x^2]))/4
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3884, 6, 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.
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx\) |
\(\Big \downarrow \) 3884 |
\(\displaystyle \int \left (\frac {a^2}{x}+\frac {2 a b \sin \left (c+d x^2\right )}{x}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x}+\frac {b^2}{2 x}\right )dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x}+\frac {2 a b \sin \left (c+d x^2\right )}{x}-\frac {b^2 \cos \left (2 c+2 d x^2\right )}{2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2 a^2+b^2\right ) \log (x)+a b \sin (c) \operatorname {CosIntegral}\left (d x^2\right )+a b \cos (c) \text {Si}\left (d x^2\right )-\frac {1}{4} b^2 \cos (2 c) \operatorname {CosIntegral}\left (2 d x^2\right )+\frac {1}{4} b^2 \sin (2 c) \text {Si}\left (2 d x^2\right )\) |
Input:
Int[(a + b*Sin[c + d*x^2])^2/x,x]
Output:
-1/4*(b^2*Cos[2*c]*CosIntegral[2*d*x^2]) + ((2*a^2 + b^2)*Log[x])/2 + a*b* CosIntegral[d*x^2]*Sin[c] + a*b*Cos[c]*SinIntegral[d*x^2] + (b^2*Sin[2*c]* SinIntegral[2*d*x^2])/4
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x _Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.94 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.12
method | result | size |
risch | \(-\frac {{\mathrm e}^{-i c} \pi \,\operatorname {csgn}\left (d \,x^{2}\right ) a b}{2}+{\mathrm e}^{-i c} \operatorname {Si}\left (d \,x^{2}\right ) a b -\frac {i {\mathrm e}^{-i c} \operatorname {expIntegral}_{1}\left (-i d \,x^{2}\right ) a b}{2}+\ln \left (x \right ) a^{2}+\frac {\ln \left (x \right ) b^{2}}{2}-\frac {i {\mathrm e}^{-2 i c} \pi \,\operatorname {csgn}\left (d \,x^{2}\right ) b^{2}}{8}+\frac {i {\mathrm e}^{-2 i c} \operatorname {Si}\left (2 d \,x^{2}\right ) b^{2}}{4}+\frac {{\mathrm e}^{-2 i c} \operatorname {expIntegral}_{1}\left (-2 i d \,x^{2}\right ) b^{2}}{8}+\frac {b^{2} {\mathrm e}^{2 i c} \operatorname {expIntegral}_{1}\left (-2 i d \,x^{2}\right )}{8}+\frac {i a b \,{\mathrm e}^{i c} \operatorname {expIntegral}_{1}\left (-i d \,x^{2}\right )}{2}\) | \(157\) |
Input:
int((a+b*sin(d*x^2+c))^2/x,x,method=_RETURNVERBOSE)
Output:
-1/2*exp(-I*c)*Pi*csgn(d*x^2)*a*b+exp(-I*c)*Si(d*x^2)*a*b-1/2*I*exp(-I*c)* Ei(1,-I*d*x^2)*a*b+ln(x)*a^2+1/2*ln(x)*b^2-1/8*I*exp(-2*I*c)*Pi*csgn(d*x^2 )*b^2+1/4*I*exp(-2*I*c)*Si(2*d*x^2)*b^2+1/8*exp(-2*I*c)*Ei(1,-2*I*d*x^2)*b ^2+1/8*b^2*exp(2*I*c)*Ei(1,-2*I*d*x^2)+1/2*I*a*b*exp(I*c)*Ei(1,-I*d*x^2)
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=-\frac {1}{4} \, b^{2} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{2}\right ) + a b \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) + \frac {1}{4} \, b^{2} \sin \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{2}\right ) + a b \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + \frac {1}{2} \, {\left (2 \, a^{2} + b^{2}\right )} \log \left (x\right ) \] Input:
integrate((a+b*sin(d*x^2+c))^2/x,x, algorithm="fricas")
Output:
-1/4*b^2*cos(2*c)*cos_integral(2*d*x^2) + a*b*cos_integral(d*x^2)*sin(c) + 1/4*b^2*sin(2*c)*sin_integral(2*d*x^2) + a*b*cos(c)*sin_integral(d*x^2) + 1/2*(2*a^2 + b^2)*log(x)
\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}{x}\, dx \] Input:
integrate((a+b*sin(d*x**2+c))**2/x,x)
Output:
Integral((a + b*sin(c + d*x**2))**2/x, x)
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=-\frac {1}{2} \, {\left ({\left (i \, {\rm Ei}\left (i \, d x^{2}\right ) - i \, {\rm Ei}\left (-i \, d x^{2}\right )\right )} \cos \left (c\right ) - {\left ({\rm Ei}\left (i \, d x^{2}\right ) + {\rm Ei}\left (-i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b - \frac {1}{8} \, {\left ({\left ({\rm Ei}\left (2 i \, d x^{2}\right ) + {\rm Ei}\left (-2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) - {\left (-i \, {\rm Ei}\left (2 i \, d x^{2}\right ) + i \, {\rm Ei}\left (-2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right ) - 4 \, \log \left (x\right )\right )} b^{2} + a^{2} \log \left (x\right ) \] Input:
integrate((a+b*sin(d*x^2+c))^2/x,x, algorithm="maxima")
Output:
-1/2*((I*Ei(I*d*x^2) - I*Ei(-I*d*x^2))*cos(c) - (Ei(I*d*x^2) + Ei(-I*d*x^2 ))*sin(c))*a*b - 1/8*((Ei(2*I*d*x^2) + Ei(-2*I*d*x^2))*cos(2*c) - (-I*Ei(2 *I*d*x^2) + I*Ei(-2*I*d*x^2))*sin(2*c) - 4*log(x))*b^2 + a^2*log(x)
Time = 0.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=-\frac {1}{4} \, b^{2} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{2}\right ) + a b \operatorname {Ci}\left (d x^{2}\right ) \sin \left (c\right ) + a b \cos \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) - \frac {1}{4} \, b^{2} \sin \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + \frac {1}{2} \, a^{2} \log \left (d x^{2}\right ) + \frac {1}{4} \, b^{2} \log \left (d x^{2}\right ) \] Input:
integrate((a+b*sin(d*x^2+c))^2/x,x, algorithm="giac")
Output:
-1/4*b^2*cos(2*c)*cos_integral(2*d*x^2) + a*b*cos_integral(d*x^2)*sin(c) + a*b*cos(c)*sin_integral(d*x^2) - 1/4*b^2*sin(2*c)*sin_integral(-2*d*x^2) + 1/2*a^2*log(d*x^2) + 1/4*b^2*log(d*x^2)
Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2}{x} \,d x \] Input:
int((a + b*sin(c + d*x^2))^2/x,x)
Output:
int((a + b*sin(c + d*x^2))^2/x, x)
\[ \int \frac {\left (a+b \sin \left (c+d x^2\right )\right )^2}{x} \, dx=\left (\int \frac {\sin \left (d \,x^{2}+c \right )^{2}}{x}d x \right ) b^{2}+2 \left (\int \frac {\sin \left (d \,x^{2}+c \right )}{x}d x \right ) a b +\mathrm {log}\left (x \right ) a^{2} \] Input:
int((a+b*sin(d*x^2+c))^2/x,x)
Output:
int(sin(c + d*x**2)**2/x,x)*b**2 + 2*int(sin(c + d*x**2)/x,x)*a*b + log(x) *a**2