Integrand size = 14, antiderivative size = 91 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=\frac {3}{8} b \cos (a) \operatorname {CosIntegral}\left (b x^2\right )-\frac {3}{8} b \cos (3 a) \operatorname {CosIntegral}\left (3 b x^2\right )-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac {3}{8} b \sin (a) \text {Si}\left (b x^2\right )+\frac {3}{8} b \sin (3 a) \text {Si}\left (3 b x^2\right ) \] Output:
3/8*b*cos(a)*Ci(b*x^2)-3/8*b*cos(3*a)*Ci(3*b*x^2)-3/8*sin(b*x^2+a)/x^2+1/8 *sin(3*b*x^2+3*a)/x^2-3/8*b*sin(a)*Si(b*x^2)+3/8*b*sin(3*a)*Si(3*b*x^2)
Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=\frac {3 b x^2 \cos (a) \operatorname {CosIntegral}\left (b x^2\right )-3 b x^2 \cos (3 a) \operatorname {CosIntegral}\left (3 b x^2\right )-3 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )-3 b x^2 \sin (a) \text {Si}\left (b x^2\right )+3 b x^2 \sin (3 a) \text {Si}\left (3 b x^2\right )}{8 x^2} \] Input:
Integrate[Sin[a + b*x^2]^3/x^3,x]
Output:
(3*b*x^2*Cos[a]*CosIntegral[b*x^2] - 3*b*x^2*Cos[3*a]*CosIntegral[3*b*x^2] - 3*Sin[a + b*x^2] + Sin[3*(a + b*x^2)] - 3*b*x^2*Sin[a]*SinIntegral[b*x^ 2] + 3*b*x^2*Sin[3*a]*SinIntegral[3*b*x^2])/(8*x^2)
Time = 0.39 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3884, 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.
\(\displaystyle \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 3884 |
\(\displaystyle \int \left (\frac {3 \sin \left (a+b x^2\right )}{4 x^3}-\frac {\sin \left (3 a+3 b x^2\right )}{4 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{8} b \cos (a) \operatorname {CosIntegral}\left (b x^2\right )-\frac {3}{8} b \cos (3 a) \operatorname {CosIntegral}\left (3 b x^2\right )-\frac {3}{8} b \sin (a) \text {Si}\left (b x^2\right )+\frac {3}{8} b \sin (3 a) \text {Si}\left (3 b x^2\right )-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2}\) |
Input:
Int[Sin[a + b*x^2]^3/x^3,x]
Output:
(3*b*Cos[a]*CosIntegral[b*x^2])/8 - (3*b*Cos[3*a]*CosIntegral[3*b*x^2])/8 - (3*Sin[a + b*x^2])/(8*x^2) + Sin[3*(a + b*x^2)]/(8*x^2) - (3*b*Sin[a]*Si nIntegral[b*x^2])/8 + (3*b*Sin[3*a]*SinIntegral[3*b*x^2])/8
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 = 2.84 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.03
method | result | size |
risch | \(-\frac {3 i \pi \,{\mathrm e}^{-3 i a} \operatorname {csgn}\left (b \,x^{2}\right ) b \,x^{2}-3 i \pi \,{\mathrm e}^{-i a} \operatorname {csgn}\left (b \,x^{2}\right ) b \,x^{2}-6 i {\mathrm e}^{-3 i a} \operatorname {Si}\left (3 b \,x^{2}\right ) b \,x^{2}+6 i {\mathrm e}^{-i a} \operatorname {Si}\left (b \,x^{2}\right ) b \,x^{2}-3 \,\operatorname {expIntegral}_{1}\left (-3 i b \,x^{2}\right ) {\mathrm e}^{-3 i a} b \,x^{2}-3 \,{\mathrm e}^{3 i a} b \,\operatorname {expIntegral}_{1}\left (-3 i b \,x^{2}\right ) x^{2}+3 \,{\mathrm e}^{i a} b \,\operatorname {expIntegral}_{1}\left (-i b \,x^{2}\right ) x^{2}+3 \,\operatorname {expIntegral}_{1}\left (-i b \,x^{2}\right ) {\mathrm e}^{-i a} b \,x^{2}+6 \sin \left (b \,x^{2}+a \right )-2 \sin \left (3 b \,x^{2}+3 a \right )}{16 x^{2}}\) | \(185\) |
Input:
int(sin(b*x^2+a)^3/x^3,x,method=_RETURNVERBOSE)
Output:
-1/16*(3*I*Pi*exp(-3*I*a)*csgn(b*x^2)*b*x^2-3*I*Pi*exp(-I*a)*csgn(b*x^2)*b *x^2-6*I*exp(-3*I*a)*Si(3*b*x^2)*b*x^2+6*I*exp(-I*a)*Si(b*x^2)*b*x^2-3*Ei( 1,-3*I*b*x^2)*exp(-3*I*a)*b*x^2-3*exp(3*I*a)*b*Ei(1,-3*I*b*x^2)*x^2+3*exp( I*a)*b*Ei(1,-I*b*x^2)*x^2+3*Ei(1,-I*b*x^2)*exp(-I*a)*b*x^2+6*sin(b*x^2+a)- 2*sin(3*b*x^2+3*a))/x^2
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=-\frac {3 \, b x^{2} \cos \left (3 \, a\right ) \operatorname {Ci}\left (3 \, b x^{2}\right ) - 3 \, b x^{2} \cos \left (a\right ) \operatorname {Ci}\left (b x^{2}\right ) - 3 \, b x^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (3 \, b x^{2}\right ) + 3 \, b x^{2} \sin \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) - 4 \, {\left (\cos \left (b x^{2} + a\right )^{2} - 1\right )} \sin \left (b x^{2} + a\right )}{8 \, x^{2}} \] Input:
integrate(sin(b*x^2+a)^3/x^3,x, algorithm="fricas")
Output:
-1/8*(3*b*x^2*cos(3*a)*cos_integral(3*b*x^2) - 3*b*x^2*cos(a)*cos_integral (b*x^2) - 3*b*x^2*sin(3*a)*sin_integral(3*b*x^2) + 3*b*x^2*sin(a)*sin_inte gral(b*x^2) - 4*(cos(b*x^2 + a)^2 - 1)*sin(b*x^2 + a))/x^2
\[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=\int \frac {\sin ^{3}{\left (a + b x^{2} \right )}}{x^{3}}\, dx \] Input:
integrate(sin(b*x**2+a)**3/x**3,x)
Output:
Integral(sin(a + b*x**2)**3/x**3, x)
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=-\frac {3}{16} \, {\left ({\left (\Gamma \left (-1, 3 i \, b x^{2}\right ) + \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) - {\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (-i \, \Gamma \left (-1, 3 i \, b x^{2}\right ) + i \, \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) + {\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} b \] Input:
integrate(sin(b*x^2+a)^3/x^3,x, algorithm="maxima")
Output:
-3/16*((gamma(-1, 3*I*b*x^2) + gamma(-1, -3*I*b*x^2))*cos(3*a) - (gamma(-1 , I*b*x^2) + gamma(-1, -I*b*x^2))*cos(a) + (-I*gamma(-1, 3*I*b*x^2) + I*ga mma(-1, -3*I*b*x^2))*sin(3*a) + (I*gamma(-1, I*b*x^2) - I*gamma(-1, -I*b*x ^2))*sin(a))*b
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (80) = 160\).
Time = 0.12 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.04 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=-\frac {3 \, {\left (b x^{2} + a\right )} b^{2} \cos \left (3 \, a\right ) \operatorname {Ci}\left (3 \, b x^{2}\right ) - 3 \, a b^{2} \cos \left (3 \, a\right ) \operatorname {Ci}\left (3 \, b x^{2}\right ) - 3 \, {\left (b x^{2} + a\right )} b^{2} \cos \left (a\right ) \operatorname {Ci}\left (b x^{2}\right ) + 3 \, a b^{2} \cos \left (a\right ) \operatorname {Ci}\left (b x^{2}\right ) + 3 \, {\left (b x^{2} + a\right )} b^{2} \sin \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) - 3 \, a b^{2} \sin \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + 3 \, {\left (b x^{2} + a\right )} b^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (-3 \, b x^{2}\right ) - 3 \, a b^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (-3 \, b x^{2}\right ) - b^{2} \sin \left (3 \, b x^{2} + 3 \, a\right ) + 3 \, b^{2} \sin \left (b x^{2} + a\right )}{8 \, b^{2} x^{2}} \] Input:
integrate(sin(b*x^2+a)^3/x^3,x, algorithm="giac")
Output:
-1/8*(3*(b*x^2 + a)*b^2*cos(3*a)*cos_integral(3*b*x^2) - 3*a*b^2*cos(3*a)* cos_integral(3*b*x^2) - 3*(b*x^2 + a)*b^2*cos(a)*cos_integral(b*x^2) + 3*a *b^2*cos(a)*cos_integral(b*x^2) + 3*(b*x^2 + a)*b^2*sin(a)*sin_integral(b* x^2) - 3*a*b^2*sin(a)*sin_integral(b*x^2) + 3*(b*x^2 + a)*b^2*sin(3*a)*sin _integral(-3*b*x^2) - 3*a*b^2*sin(3*a)*sin_integral(-3*b*x^2) - b^2*sin(3* b*x^2 + 3*a) + 3*b^2*sin(b*x^2 + a))/(b^2*x^2)
Timed out. \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx=\int \frac {{\sin \left (b\,x^2+a\right )}^3}{x^3} \,d x \] Input:
int(sin(a + b*x^2)^3/x^3,x)
Output:
int(sin(a + b*x^2)^3/x^3, x)
\[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx =\text {Too large to display} \] Input:
int(sin(b*x^2+a)^3/x^3,x)
Output:
( - 3*cos(a + b*x**2)*sin(a + b*x**2)*tan((a + b*x**2)/2)**6 - 9*cos(a + b *x**2)*sin(a + b*x**2)*tan((a + b*x**2)/2)**4 - 9*cos(a + b*x**2)*sin(a + b*x**2)*tan((a + b*x**2)/2)**2 - 3*cos(a + b*x**2)*sin(a + b*x**2) + 40*in t(tan((a + b*x**2)/2)**3/(tan((a + b*x**2)/2)**6*x**3 + 3*tan((a + b*x**2) /2)**4*x**3 + 3*tan((a + b*x**2)/2)**2*x**3 + x**3),x)*tan((a + b*x**2)/2) **6*x**2 + 120*int(tan((a + b*x**2)/2)**3/(tan((a + b*x**2)/2)**6*x**3 + 3 *tan((a + b*x**2)/2)**4*x**3 + 3*tan((a + b*x**2)/2)**2*x**3 + x**3),x)*ta n((a + b*x**2)/2)**4*x**2 + 120*int(tan((a + b*x**2)/2)**3/(tan((a + b*x** 2)/2)**6*x**3 + 3*tan((a + b*x**2)/2)**4*x**3 + 3*tan((a + b*x**2)/2)**2*x **3 + x**3),x)*tan((a + b*x**2)/2)**2*x**2 + 40*int(tan((a + b*x**2)/2)**3 /(tan((a + b*x**2)/2)**6*x**3 + 3*tan((a + b*x**2)/2)**4*x**3 + 3*tan((a + b*x**2)/2)**2*x**3 + x**3),x)*x**2 - sin(a + b*x**2)**3*tan((a + b*x**2)/ 2)**6 - 3*sin(a + b*x**2)**3*tan((a + b*x**2)/2)**4 - 3*sin(a + b*x**2)**3 *tan((a + b*x**2)/2)**2 - sin(a + b*x**2)**3 - 3*sin(a + b*x**2)*tan((a + b*x**2)/2)**6 - 9*sin(a + b*x**2)*tan((a + b*x**2)/2)**4 - 9*sin(a + b*x** 2)*tan((a + b*x**2)/2)**2 - 3*sin(a + b*x**2) + 20*tan((a + b*x**2)/2)**3 + 12*tan((a + b*x**2)/2))/(5*x**2*(tan((a + b*x**2)/2)**6 + 3*tan((a + b*x **2)/2)**4 + 3*tan((a + b*x**2)/2)**2 + 1))