Integrand size = 18, antiderivative size = 337 \[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\frac {2 b f^2 (d e-c f) (c+d x) \cos \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {b f^3 (c+d x)^2 \cos \left (\frac {b}{(c+d x)^2}\right )}{4 d^4}-\frac {3 b f (d e-c f)^2 \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )}{2 d^4}-\frac {\sqrt {b} (d e-c f)^3 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^4}+\frac {2 b^{3/2} f^2 (d e-c f) \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^4}+\frac {(d e-c f)^3 (c+d x) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {3 f (d e-c f)^2 (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{2 d^4}+\frac {f^2 (d e-c f) (c+d x)^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {f^3 (c+d x)^4 \sin \left (\frac {b}{(c+d x)^2}\right )}{4 d^4}+\frac {b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )}{4 d^4} \] Output:
2*b*f^2*(-c*f+d*e)*(d*x+c)*cos(b/(d*x+c)^2)/d^4+1/4*b*f^3*(d*x+c)^2*cos(b/ (d*x+c)^2)/d^4-3/2*b*f*(-c*f+d*e)^2*Ci(b/(d*x+c)^2)/d^4-b^(1/2)*(-c*f+d*e) ^3*2^(1/2)*Pi^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))/d^4+2*b^(3/ 2)*f^2*(-c*f+d*e)*2^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+ c))/d^4+(-c*f+d*e)^3*(d*x+c)*sin(b/(d*x+c)^2)/d^4+3/2*f*(-c*f+d*e)^2*(d*x+ c)^2*sin(b/(d*x+c)^2)/d^4+f^2*(-c*f+d*e)*(d*x+c)^3*sin(b/(d*x+c)^2)/d^4+1/ 4*f^3*(d*x+c)^4*sin(b/(d*x+c)^2)/d^4+1/4*b^2*f^3*Si(b/(d*x+c)^2)/d^4
Time = 6.70 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.31 \[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\frac {8 b c d e f^2 \cos \left (\frac {b}{(c+d x)^2}\right )-7 b c^2 f^3 \cos \left (\frac {b}{(c+d x)^2}\right )+8 b d^2 e f^2 x \cos \left (\frac {b}{(c+d x)^2}\right )-6 b c d f^3 x \cos \left (\frac {b}{(c+d x)^2}\right )+b d^2 f^3 x^2 \cos \left (\frac {b}{(c+d x)^2}\right )-6 b f (d e-c f)^2 \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )-4 \sqrt {b} (d e-c f)^3 \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+8 b^{3/2} d e f^2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )-8 b^{3/2} c f^3 \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+4 c d^3 e^3 \sin \left (\frac {b}{(c+d x)^2}\right )-6 c^2 d^2 e^2 f \sin \left (\frac {b}{(c+d x)^2}\right )+4 c^3 d e f^2 \sin \left (\frac {b}{(c+d x)^2}\right )-c^4 f^3 \sin \left (\frac {b}{(c+d x)^2}\right )+4 d^4 e^3 x \sin \left (\frac {b}{(c+d x)^2}\right )+6 d^4 e^2 f x^2 \sin \left (\frac {b}{(c+d x)^2}\right )+4 d^4 e f^2 x^3 \sin \left (\frac {b}{(c+d x)^2}\right )+d^4 f^3 x^4 \sin \left (\frac {b}{(c+d x)^2}\right )+b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )}{4 d^4} \] Input:
Integrate[(e + f*x)^3*Sin[b/(c + d*x)^2],x]
Output:
(8*b*c*d*e*f^2*Cos[b/(c + d*x)^2] - 7*b*c^2*f^3*Cos[b/(c + d*x)^2] + 8*b*d ^2*e*f^2*x*Cos[b/(c + d*x)^2] - 6*b*c*d*f^3*x*Cos[b/(c + d*x)^2] + b*d^2*f ^3*x^2*Cos[b/(c + d*x)^2] - 6*b*f*(d*e - c*f)^2*CosIntegral[b/(c + d*x)^2] - 4*Sqrt[b]*(d*e - c*f)^3*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d *x)] + 8*b^(3/2)*d*e*f^2*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x )] - 8*b^(3/2)*c*f^3*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 4*c*d^3*e^3*Sin[b/(c + d*x)^2] - 6*c^2*d^2*e^2*f*Sin[b/(c + d*x)^2] + 4*c ^3*d*e*f^2*Sin[b/(c + d*x)^2] - c^4*f^3*Sin[b/(c + d*x)^2] + 4*d^4*e^3*x*S in[b/(c + d*x)^2] + 6*d^4*e^2*f*x^2*Sin[b/(c + d*x)^2] + 4*d^4*e*f^2*x^3*S in[b/(c + d*x)^2] + d^4*f^3*x^4*Sin[b/(c + d*x)^2] + b^2*f^3*SinIntegral[b /(c + d*x)^2])/(4*d^4)
Time = 0.54 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
| \(\displaystyle \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx\) |
| \(\Big \downarrow \) 3914 |
| \(\displaystyle \frac {\int \left (\sin \left (\frac {b}{(c+d x)^2}\right ) (d e-c f)^3+3 f (c+d x) \sin \left (\frac {b}{(c+d x)^2}\right ) (d e-c f)^2+3 f^2 (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right ) (d e-c f)+f^3 (c+d x)^3 \sin \left (\frac {b}{(c+d x)^2}\right )\right )d(c+d x)}{d^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \sqrt {2 \pi } b^{3/2} f^2 (d e-c f) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+\frac {1}{4} b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )-\frac {3}{2} b f (d e-c f)^2 \operatorname {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )+f^2 (c+d x)^3 (d e-c f) \sin \left (\frac {b}{(c+d x)^2}\right )+2 b f^2 (c+d x) (d e-c f) \cos \left (\frac {b}{(c+d x)^2}\right )-\sqrt {2 \pi } \sqrt {b} (d e-c f)^3 \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+\frac {3}{2} f (c+d x)^2 (d e-c f)^2 \sin \left (\frac {b}{(c+d x)^2}\right )+(c+d x) (d e-c f)^3 \sin \left (\frac {b}{(c+d x)^2}\right )+\frac {1}{4} f^3 (c+d x)^4 \sin \left (\frac {b}{(c+d x)^2}\right )+\frac {1}{4} b f^3 (c+d x)^2 \cos \left (\frac {b}{(c+d x)^2}\right )}{d^4}\) |
Input:
Int[(e + f*x)^3*Sin[b/(c + d*x)^2],x]
Output:
(2*b*f^2*(d*e - c*f)*(c + d*x)*Cos[b/(c + d*x)^2] + (b*f^3*(c + d*x)^2*Cos [b/(c + d*x)^2])/4 - (3*b*f*(d*e - c*f)^2*CosIntegral[b/(c + d*x)^2])/2 - Sqrt[b]*(d*e - c*f)^3*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 2*b^(3/2)*f^2*(d*e - c*f)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + (d*e - c*f)^3*(c + d*x)*Sin[b/(c + d*x)^2] + (3*f*(d*e - c*f)^2*(c + d*x)^2*Sin[b/(c + d*x)^2])/2 + f^2*(d*e - c*f)*(c + d*x)^3*Sin[b/(c + d *x)^2] + (f^3*(c + d*x)^4*Sin[b/(c + d*x)^2])/4 + (b^2*f^3*SinIntegral[b/( c + d*x)^2])/4)/d^4
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]
Time = 2.18 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-\left (c f -d e \right )^{3} \left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\left (c f -d e \right )^{3} \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )+\frac {3 f \left (c f -d e \right )^{2} \left (d x +c \right )^{2} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {3 f \left (c f -d e \right )^{2} b \,\operatorname {Ci}\left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-f^{2} \left (c f -d e \right ) \left (d x +c \right )^{3} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+2 f^{2} \left (c f -d e \right ) b \left (-\left (d x +c \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )+\frac {f^{3} \left (d x +c \right )^{4} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{4}-\frac {f^{3} b \left (-\frac {\left (d x +c \right )^{2} \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {b \,\operatorname {Si}\left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}\right )}{2}}{d^{4}}\) | \(277\) |
| default | \(\frac {-\left (c f -d e \right )^{3} \left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\left (c f -d e \right )^{3} \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )+\frac {3 f \left (c f -d e \right )^{2} \left (d x +c \right )^{2} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {3 f \left (c f -d e \right )^{2} b \,\operatorname {Ci}\left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-f^{2} \left (c f -d e \right ) \left (d x +c \right )^{3} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+2 f^{2} \left (c f -d e \right ) b \left (-\left (d x +c \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )+\frac {f^{3} \left (d x +c \right )^{4} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{4}-\frac {f^{3} b \left (-\frac {\left (d x +c \right )^{2} \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {b \,\operatorname {Si}\left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}\right )}{2}}{d^{4}}\) | \(277\) |
| risch | \(\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right ) c^{3} f^{3}}{2 d^{4} \sqrt {-i b}}-\frac {3 b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right ) c^{2} e \,f^{2}}{2 d^{3} \sqrt {-i b}}+\frac {3 b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right ) c \,e^{2} f}{2 d^{2} \sqrt {-i b}}-\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right ) e^{3}}{2 d \sqrt {-i b}}+\frac {3 b \,\operatorname {expIntegral}_{1}\left (-\frac {i b}{\left (d x +c \right )^{2}}\right ) c^{2} f^{3}}{4 d^{4}}-\frac {3 b \,\operatorname {expIntegral}_{1}\left (-\frac {i b}{\left (d x +c \right )^{2}}\right ) c e \,f^{2}}{2 d^{3}}+\frac {3 b \,\operatorname {expIntegral}_{1}\left (-\frac {i b}{\left (d x +c \right )^{2}}\right ) e^{2} f}{4 d^{2}}-\frac {i b^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) c \,f^{3}}{d^{4} \sqrt {i b}}+\frac {i b^{2} f^{3} c \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right )}{d^{4} \sqrt {-i b}}-\frac {i b^{2} \operatorname {expIntegral}_{1}\left (\frac {i b}{\left (d x +c \right )^{2}}\right ) f^{3}}{8 d^{4}}+\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) c^{3} f^{3}}{2 d^{4} \sqrt {i b}}-\frac {3 b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) c^{2} e \,f^{2}}{2 d^{3} \sqrt {i b}}+\frac {3 b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) c \,e^{2} f}{2 d^{2} \sqrt {i b}}-\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) e^{3}}{2 d \sqrt {i b}}+\frac {3 b \,\operatorname {expIntegral}_{1}\left (\frac {i b}{\left (d x +c \right )^{2}}\right ) c^{2} f^{3}}{4 d^{4}}-\frac {3 b \,\operatorname {expIntegral}_{1}\left (\frac {i b}{\left (d x +c \right )^{2}}\right ) c e \,f^{2}}{2 d^{3}}+\frac {3 b \,\operatorname {expIntegral}_{1}\left (\frac {i b}{\left (d x +c \right )^{2}}\right ) e^{2} f}{4 d^{2}}-\frac {i b^{2} e \,f^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right )}{d^{3} \sqrt {-i b}}+\frac {i b^{2} f^{3} \operatorname {expIntegral}_{1}\left (-\frac {i b}{\left (d x +c \right )^{2}}\right )}{8 d^{4}}+\frac {i b^{2} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) e \,f^{2}}{d^{3} \sqrt {i b}}-\frac {b \,f^{2} \left (-d^{2} f \,x^{2}+6 c d f x -8 d^{2} e x +7 c^{2} f -8 c d e \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{4 d^{4}}-\frac {\left (-f^{3} x^{4} d^{4}-4 x^{3} d^{4} e \,f^{2}-6 x^{2} d^{4} e^{2} f -4 x \,d^{4} e^{3}+c^{4} f^{3}-4 c^{3} d e \,f^{2}+6 c^{2} d^{2} e^{2} f -4 c \,d^{3} e^{3}\right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{4 d^{4}}\) | \(777\) |
| parts | \(\text {Expression too large to display}\) | \(828\) |
Input:
int((f*x+e)^3*sin(b/(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d^4*(-(c*f-d*e)^3*(d*x+c)*sin(b/(d*x+c)^2)+(c*f-d*e)^3*b^(1/2)*2^(1/2)*P i^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))+3/2*f*(c*f-d*e)^2*(d*x+ c)^2*sin(b/(d*x+c)^2)-3/2*f*(c*f-d*e)^2*b*Ci(b/(d*x+c)^2)-f^2*(c*f-d*e)*(d *x+c)^3*sin(b/(d*x+c)^2)+2*f^2*(c*f-d*e)*b*(-(d*x+c)*cos(b/(d*x+c)^2)-b^(1 /2)*2^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))+1/4*f^3*( d*x+c)^4*sin(b/(d*x+c)^2)-1/2*f^3*b*(-1/2*(d*x+c)^2*cos(b/(d*x+c)^2)-1/2*b *Si(b/(d*x+c)^2)))
Time = 0.11 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.18 \[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\frac {b^{2} f^{3} \operatorname {Si}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, \sqrt {2} \pi {\left (d^{4} e^{3} - 3 \, c d^{3} e^{2} f + 3 \, c^{2} d^{2} e f^{2} - c^{3} d f^{3}\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) + 8 \, \sqrt {2} \pi {\left (b d^{2} e f^{2} - b c d f^{3}\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {S}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) + {\left (b d^{2} f^{3} x^{2} + 8 \, b c d e f^{2} - 7 \, b c^{2} f^{3} + 2 \, {\left (4 \, b d^{2} e f^{2} - 3 \, b c d f^{3}\right )} x\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 6 \, {\left (b d^{2} e^{2} f - 2 \, b c d e f^{2} + b c^{2} f^{3}\right )} \operatorname {Ci}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + {\left (d^{4} f^{3} x^{4} + 4 \, d^{4} e f^{2} x^{3} + 6 \, d^{4} e^{2} f x^{2} + 4 \, d^{4} e^{3} x + 4 \, c d^{3} e^{3} - 6 \, c^{2} d^{2} e^{2} f + 4 \, c^{3} d e f^{2} - c^{4} f^{3}\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{4 \, d^{4}} \] Input:
integrate((f*x+e)^3*sin(b/(d*x+c)^2),x, algorithm="fricas")
Output:
1/4*(b^2*f^3*sin_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) - 4*sqrt(2)*pi*(d^4 *e^3 - 3*c*d^3*e^2*f + 3*c^2*d^2*e*f^2 - c^3*d*f^3)*sqrt(b/(pi*d^2))*fresn el_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + 8*sqrt(2)*pi*(b*d^2*e*f^2 - b*c*d*f^3)*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + (b*d^2*f^3*x^2 + 8*b*c*d*e*f^2 - 7*b*c^2*f^3 + 2*(4*b*d^2*e*f^2 - 3 *b*c*d*f^3)*x)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - 6*(b*d^2*e^2*f - 2*b*c*d *e*f^2 + b*c^2*f^3)*cos_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) + (d^4*f^3*x ^4 + 4*d^4*e*f^2*x^3 + 6*d^4*e^2*f*x^2 + 4*d^4*e^3*x + 4*c*d^3*e^3 - 6*c^2 *d^2*e^2*f + 4*c^3*d*e*f^2 - c^4*f^3)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^ 4
\[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\int \left (e + f x\right )^{3} \sin {\left (\frac {b}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}\, dx \] Input:
integrate((f*x+e)**3*sin(b/(d*x+c)**2),x)
Output:
Integral((e + f*x)**3*sin(b/(c**2 + 2*c*d*x + d**2*x**2)), x)
\[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\int { {\left (f x + e\right )}^{3} \sin \left (\frac {b}{{\left (d x + c\right )}^{2}}\right ) \,d x } \] Input:
integrate((f*x+e)^3*sin(b/(d*x+c)^2),x, algorithm="maxima")
Output:
-1/4*(4*d^3*integrate(1/4*((4*b*c^3*d*e*f^2 - 3*b*c^4*f^3 - 6*(b*d^4*e^2*f - 2*b*c*d^3*e*f^2 + b*c^2*d^2*f^3)*x^2 - 4*(b*d^4*e^3 - 3*b*c^2*d^2*e*f^2 + 2*b*c^3*d*f^3)*x)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) + (b^2*d^2*f^3*x^2 + 2*(4*b^2*d^2*e*f^2 - 3*b^2*c*d*f^3)*x)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/ (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3), x) + 4*d^3*integrate(1/4* ((4*b*c^3*d*e*f^2 - 3*b*c^4*f^3 - 6*(b*d^4*e^2*f - 2*b*c*d^3*e*f^2 + b*c^2 *d^2*f^3)*x^2 - 4*(b*d^4*e^3 - 3*b*c^2*d^2*e*f^2 + 2*b*c^3*d*f^3)*x)*cos(b /(d^2*x^2 + 2*c*d*x + c^2)) + (b^2*d^2*f^3*x^2 + 2*(4*b^2*d^2*e*f^2 - 3*b^ 2*c*d*f^3)*x)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/((d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cos(b/(d^2*x^2 + 2*c*d*x + c^2))^2 + (d^6*x^3 + 3*c *d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*sin(b/(d^2*x^2 + 2*c*d*x + c^2))^2), x) - (b*d*f^3*x^2 + 2*(4*b*d*e*f^2 - 3*b*c*f^3)*x)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - (d^3*f^3*x^4 + 4*d^3*e*f^2*x^3 + 6*d^3*e^2*f*x^2 + 4*d^3*e^3*x)*s in(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^3
\[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\int { {\left (f x + e\right )}^{3} \sin \left (\frac {b}{{\left (d x + c\right )}^{2}}\right ) \,d x } \] Input:
integrate((f*x+e)^3*sin(b/(d*x+c)^2),x, algorithm="giac")
Output:
integrate((f*x + e)^3*sin(b/(d*x + c)^2), x)
Timed out. \[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\int \sin \left (\frac {b}{{\left (c+d\,x\right )}^2}\right )\,{\left (e+f\,x\right )}^3 \,d x \] Input:
int(sin(b/(c + d*x)^2)*(e + f*x)^3,x)
Output:
int(sin(b/(c + d*x)^2)*(e + f*x)^3, x)
\[ \int (e+f x)^3 \sin \left (\frac {b}{(c+d x)^2}\right ) \, dx=\text {too large to display} \] Input:
int((f*x+e)^3*sin(b/(d*x+c)^2),x)
Output:
( - 120*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**6*f**3 - 120*cos(b/(c* *2 + 2*c*d*x + d**2*x**2))*b**2*c**5*d*e*f**2 - 456*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**5*d*f**3*x - 456*cos(b/(c**2 + 2*c*d*x + d**2*x**2)) *b**2*c**4*d**2*e*f**2*x - 624*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c* *4*d**2*f**3*x**2 - 624*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**3*d**3 *e*f**2*x**2 - 336*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**3*d**3*f**3 *x**3 - 336*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**2*d**4*e*f**2*x**3 - 24*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c**2*d**4*f**3*x**4 - 24*co s(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c*d**5*e*f**2*x**4 + 24*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*c*d**5*f**3*x**5 + 24*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*b**2*d**6*e*f**2*x**5 - 2250*cos(b/(c**2 + 2*c*d*x + d**2*x* *2))*c**10*f**3 - 1980*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**9*d*e*f**2 - 9000*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**9*d*f**3*x + 630*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**8*d**2*e**2*f - 7920*cos(b/(c**2 + 2*c*d*x + d* *2*x**2))*c**8*d**2*e*f**2*x - 13500*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c **8*d**2*f**3*x**2 + 240*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**7*d**3*e** 3 + 2520*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**7*d**3*e**2*f*x - 11880*co s(b/(c**2 + 2*c*d*x + d**2*x**2))*c**7*d**3*e*f**2*x**2 - 9000*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**7*d**3*f**3*x**3 + 960*cos(b/(c**2 + 2*c*d*x + d**2*x**2))*c**6*d**4*e**3*x + 3780*cos(b/(c**2 + 2*c*d*x + d**2*x**2)...