Integrand size = 20, antiderivative size = 330 \[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=-\frac {b f^2 \cos (a) \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3}-\frac {i e^{i a} f (d e-c f) \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2 \Gamma \left (-\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )}{3 d^3}+\frac {i e^{-i a} f (d e-c f) \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (c+d x)^2 \Gamma \left (-\frac {2}{3},\frac {i b}{(c+d x)^3}\right )}{3 d^3}-\frac {i e^{i a} (d e-c f)^2 \sqrt [3]{-\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{6 d^3}+\frac {i e^{-i a} (d e-c f)^2 \sqrt [3]{\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (-\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{6 d^3}+\frac {f^2 (c+d x)^3 \sin \left (a+\frac {b}{(c+d x)^3}\right )}{3 d^3}+\frac {b f^2 \sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )}{3 d^3} \]
-1/3*b*f^2*Ci(b/(d*x+c)^3)*cos(a)/d^3-1/3*I*exp(I*a)*f*(-c*f+d*e)*(-I*b/(d *x+c)^3)^(2/3)*(d*x+c)^2*GAMMA(-2/3,-I*b/(d*x+c)^3)/d^3+1/3*I*f*(-c*f+d*e) *(I*b/(d*x+c)^3)^(2/3)*(d*x+c)^2*GAMMA(-2/3,I*b/(d*x+c)^3)/d^3/exp(I*a)-1/ 6*I*exp(I*a)*(-c*f+d*e)^2*(-I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GAMMA(-1/3,-I*b/( d*x+c)^3)/d^3+1/6*I*(-c*f+d*e)^2*(I*b/(d*x+c)^3)^(1/3)*(d*x+c)*GAMMA(-1/3, I*b/(d*x+c)^3)/d^3/exp(I*a)+1/3*b*f^2*Si(b/(d*x+c)^3)*sin(a)/d^3+1/3*f^2*( d*x+c)^3*sin(a+b/(d*x+c)^3)/d^3
Time = 2.57 (sec) , antiderivative size = 620, normalized size of antiderivative = 1.88 \[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=\frac {\frac {3 b d e f \cos (a) \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{-\frac {i b}{(c+d x)^3}}}+\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{\frac {i b}{(c+d x)^3}}}\right )}{c+d x}-\frac {3 b c f^2 \cos (a) \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{-\frac {i b}{(c+d x)^3}}}+\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{\frac {i b}{(c+d x)^3}}}\right )}{c+d x}-3 i (d e-c f)^2 \sqrt [3]{\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (\frac {2}{3},\frac {i b}{(c+d x)^3}\right ) (\cos (a)-i \sin (a))+3 i (d e-c f)^2 \sqrt [3]{-\frac {i b}{(c+d x)^3}} (c+d x) \Gamma \left (\frac {2}{3},-\frac {i b}{(c+d x)^3}\right ) (\cos (a)+i \sin (a))+2 (c+d x) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right ) \cos \left (\frac {b}{(c+d x)^3}\right ) \sin (a)+\frac {3 i b d e f \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{-\frac {i b}{(c+d x)^3}}}-\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{\frac {i b}{(c+d x)^3}}}\right ) \sin (a)}{c+d x}-\frac {3 i b c f^2 \left (\frac {\Gamma \left (\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{-\frac {i b}{(c+d x)^3}}}-\frac {\Gamma \left (\frac {1}{3},\frac {i b}{(c+d x)^3}\right )}{\sqrt [3]{\frac {i b}{(c+d x)^3}}}\right ) \sin (a)}{c+d x}+2 (c+d x) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right ) \cos (a) \sin \left (\frac {b}{(c+d x)^3}\right )-2 b f^2 \left (\cos (a) \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^3}\right )-\sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )\right )}{6 d^3} \]
((3*b*d*e*f*Cos[a]*(Gamma[1/3, ((-I)*b)/(c + d*x)^3]/(((-I)*b)/(c + d*x)^3 )^(1/3) + Gamma[1/3, (I*b)/(c + d*x)^3]/((I*b)/(c + d*x)^3)^(1/3)))/(c + d *x) - (3*b*c*f^2*Cos[a]*(Gamma[1/3, ((-I)*b)/(c + d*x)^3]/(((-I)*b)/(c + d *x)^3)^(1/3) + Gamma[1/3, (I*b)/(c + d*x)^3]/((I*b)/(c + d*x)^3)^(1/3)))/( c + d*x) - (3*I)*(d*e - c*f)^2*((I*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[2 /3, (I*b)/(c + d*x)^3]*(Cos[a] - I*Sin[a]) + (3*I)*(d*e - c*f)^2*(((-I)*b) /(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[2/3, ((-I)*b)/(c + d*x)^3]*(Cos[a] + I *Sin[a]) + 2*(c + d*x)*(c^2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[b/(c + d*x)^3]*Sin[a] + ((3*I)*b*d*e*f*(Gamma[1/3, ((-I)* b)/(c + d*x)^3]/(((-I)*b)/(c + d*x)^3)^(1/3) - Gamma[1/3, (I*b)/(c + d*x)^ 3]/((I*b)/(c + d*x)^3)^(1/3))*Sin[a])/(c + d*x) - ((3*I)*b*c*f^2*(Gamma[1/ 3, ((-I)*b)/(c + d*x)^3]/(((-I)*b)/(c + d*x)^3)^(1/3) - Gamma[1/3, (I*b)/( c + d*x)^3]/((I*b)/(c + d*x)^3)^(1/3))*Sin[a])/(c + d*x) + 2*(c + d*x)*(c^ 2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[a]*Sin[b/ (c + d*x)^3] - 2*b*f^2*(Cos[a]*CosIntegral[b/(c + d*x)^3] - Sin[a]*SinInte gral[b/(c + d*x)^3]))/(6*d^3)
Time = 0.44 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3914, 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.
| \(\displaystyle \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx\) |
| \(\Big \downarrow \) 3914 |
| \(\displaystyle \frac {\int \left (\sin \left (a+\frac {b}{(c+d x)^3}\right ) (d e-c f)^2+2 f (c+d x) \sin \left (a+\frac {b}{(c+d x)^3}\right ) (d e-c f)+f^2 (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right )\right )d(c+d x)}{d^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{3} b f^2 \cos (a) \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^3}\right )-\frac {1}{3} i e^{i a} f (c+d x)^2 \left (-\frac {i b}{(c+d x)^3}\right )^{2/3} (d e-c f) \Gamma \left (-\frac {2}{3},-\frac {i b}{(c+d x)^3}\right )+\frac {1}{3} i e^{-i a} f (c+d x)^2 \left (\frac {i b}{(c+d x)^3}\right )^{2/3} (d e-c f) \Gamma \left (-\frac {2}{3},\frac {i b}{(c+d x)^3}\right )-\frac {1}{6} i e^{i a} (c+d x) \sqrt [3]{-\frac {i b}{(c+d x)^3}} (d e-c f)^2 \Gamma \left (-\frac {1}{3},-\frac {i b}{(c+d x)^3}\right )+\frac {1}{6} i e^{-i a} (c+d x) \sqrt [3]{\frac {i b}{(c+d x)^3}} (d e-c f)^2 \Gamma \left (-\frac {1}{3},\frac {i b}{(c+d x)^3}\right )+\frac {1}{3} b f^2 \sin (a) \text {Si}\left (\frac {b}{(c+d x)^3}\right )+\frac {1}{3} f^2 (c+d x)^3 \sin \left (a+\frac {b}{(c+d x)^3}\right )}{d^3}\) |
(-1/3*(b*f^2*Cos[a]*CosIntegral[b/(c + d*x)^3]) - (I/3)*E^(I*a)*f*(d*e - c *f)*(((-I)*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2/3, ((-I)*b)/(c + d*x )^3] + ((I/3)*f*(d*e - c*f)*((I*b)/(c + d*x)^3)^(2/3)*(c + d*x)^2*Gamma[-2 /3, (I*b)/(c + d*x)^3])/E^(I*a) - (I/6)*E^(I*a)*(d*e - c*f)^2*(((-I)*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, ((-I)*b)/(c + d*x)^3] + ((I/6)*(d*e - c*f)^2*((I*b)/(c + d*x)^3)^(1/3)*(c + d*x)*Gamma[-1/3, (I*b)/(c + d*x)^ 3])/E^(I*a) + (f^2*(c + d*x)^3*Sin[a + b/(c + d*x)^3])/3 + (b*f^2*Sin[a]*S inIntegral[b/(c + d*x)^3])/3)/d^3
3.2.82.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f _.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat or[n], 1]}, Simp[k/f^(m + 1) Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x ^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x ]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
\[\int \left (f x +e \right )^{2} \sin \left (a +\frac {b}{\left (d x +c \right )^{3}}\right )d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (264) = 528\).
Time = 0.11 (sec) , antiderivative size = 576, normalized size of antiderivative = 1.75 \[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=-\frac {2 \, b f^{2} \cos \left (a\right ) \operatorname {Ci}\left (\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - 2 \, b f^{2} \sin \left (a\right ) \operatorname {Si}\left (\frac {b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 3 \, {\left ({\left (i \, d^{3} e f - i \, c d^{2} f^{2}\right )} \cos \left (a\right ) + {\left (d^{3} e f - c d^{2} f^{2}\right )} \sin \left (a\right )\right )} \left (\frac {i \, b}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, \frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 3 \, {\left ({\left (-i \, d^{3} e f + i \, c d^{2} f^{2}\right )} \cos \left (a\right ) + {\left (d^{3} e f - c d^{2} f^{2}\right )} \sin \left (a\right )\right )} \left (-\frac {i \, b}{d^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 3 \, {\left ({\left (i \, d^{3} e^{2} - 2 i \, c d^{2} e f + i \, c^{2} d f^{2}\right )} \cos \left (a\right ) + {\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \sin \left (a\right )\right )} \left (\frac {i \, b}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, \frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 3 \, {\left ({\left (-i \, d^{3} e^{2} + 2 i \, c d^{2} e f - i \, c^{2} d f^{2}\right )} \cos \left (a\right ) + {\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \sin \left (a\right )\right )} \left (-\frac {i \, b}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {i \, b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - 2 \, {\left (d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 3 \, c d^{2} e^{2} - 3 \, c^{2} d e f + c^{3} f^{2}\right )} \sin \left (\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}{6 \, d^{3}} \]
-1/6*(2*b*f^2*cos(a)*cos_integral(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c ^3)) - 2*b*f^2*sin(a)*sin_integral(b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + 3*((I*d^3*e*f - I*c*d^2*f^2)*cos(a) + (d^3*e*f - c*d^2*f^2)*sin(a) )*(I*b/d^3)^(2/3)*gamma(1/3, I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3) ) + 3*((-I*d^3*e*f + I*c*d^2*f^2)*cos(a) + (d^3*e*f - c*d^2*f^2)*sin(a))*( -I*b/d^3)^(2/3)*gamma(1/3, -I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + 3*((I*d^3*e^2 - 2*I*c*d^2*e*f + I*c^2*d*f^2)*cos(a) + (d^3*e^2 - 2*c*d^ 2*e*f + c^2*d*f^2)*sin(a))*(I*b/d^3)^(1/3)*gamma(2/3, I*b/(d^3*x^3 + 3*c*d ^2*x^2 + 3*c^2*d*x + c^3)) + 3*((-I*d^3*e^2 + 2*I*c*d^2*e*f - I*c^2*d*f^2) *cos(a) + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*sin(a))*(-I*b/d^3)^(1/3)*gam ma(2/3, -I*b/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) - 2*(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*d^3*e^2*x + 3*c*d^2*e^2 - 3*c^2*d*e*f + c^3*f^2)*sin((a *d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)))/d^3
\[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=\int \left (e + f x\right )^{2} \sin {\left (a + \frac {b}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}} \right )}\, dx \]
\[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=\int { {\left (f x + e\right )}^{2} \sin \left (a + \frac {b}{{\left (d x + c\right )}^{3}}\right ) \,d x } \]
1/3*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*x)*sin((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c ^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + integrate (1/2*(b*d*f^2*x^3 + 3*b*d*e*f*x^2 + 3*b*d*e^2*x)*cos((a*d^3*x^3 + 3*a*c*d^ 2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3) )/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x) + integrat e(1/2*(b*d*f^2*x^3 + 3*b*d*e*f*x^2 + 3*b*d*e^2*x)*cos((a*d^3*x^3 + 3*a*c*d ^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3 ))/((d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4)*cos((a*d^3*x ^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c ^2*d*x + c^3))^2 + (d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^ 4)*sin((a*d^3*x^3 + 3*a*c*d^2*x^2 + 3*a*c^2*d*x + a*c^3 + b)/(d^3*x^3 + 3* c*d^2*x^2 + 3*c^2*d*x + c^3))^2), x)
\[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=\int { {\left (f x + e\right )}^{2} \sin \left (a + \frac {b}{{\left (d x + c\right )}^{3}}\right ) \,d x } \]
Timed out. \[ \int (e+f x)^2 \sin \left (a+\frac {b}{(c+d x)^3}\right ) \, dx=\int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^3}\right )\,{\left (e+f\,x\right )}^2 \,d x \]