Integrand size = 22, antiderivative size = 513 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\frac {6 f (d e-c f) \cos \left (a+b (c+d x)^{2/3}\right )}{b^3 d^3}-\frac {3 (d e-c f)^2 \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^3}+\frac {105 f^2 (c+d x) \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d^3}-\frac {3 f (d e-c f) (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{b d^3}-\frac {3 f^2 (c+d x)^{7/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^3}+\frac {3 (d e-c f)^2 \sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^3}+\frac {315 f^2 \sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{16 b^{9/2} d^3}+\frac {315 f^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{16 b^{9/2} d^3}-\frac {3 (d e-c f)^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{2 b^{3/2} d^3}-\frac {315 f^2 \sqrt [3]{c+d x} \sin \left (a+b (c+d x)^{2/3}\right )}{16 b^4 d^3}+\frac {6 f (d e-c f) (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{b^2 d^3}+\frac {21 f^2 (c+d x)^{5/3} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d^3} \] Output:
6*f*(-c*f+d*e)*cos(a+b*(d*x+c)^(2/3))/b^3/d^3-3/2*(-c*f+d*e)^2*(d*x+c)^(1/ 3)*cos(a+b*(d*x+c)^(2/3))/b/d^3+105/8*f^2*(d*x+c)*cos(a+b*(d*x+c)^(2/3))/b ^3/d^3-3*f*(-c*f+d*e)*(d*x+c)^(4/3)*cos(a+b*(d*x+c)^(2/3))/b/d^3-3/2*f^2*( d*x+c)^(7/3)*cos(a+b*(d*x+c)^(2/3))/b/d^3+3/4*(-c*f+d*e)^2*2^(1/2)*Pi^(1/2 )*cos(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))/b^(3/2)/d^3+315/ 32*f^2*2^(1/2)*Pi^(1/2)*cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^( 1/3))/b^(9/2)/d^3+315/32*f^2*2^(1/2)*Pi^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^ (1/2)*(d*x+c)^(1/3))*sin(a)/b^(9/2)/d^3-3/4*(-c*f+d*e)^2*2^(1/2)*Pi^(1/2)* FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))*sin(a)/b^(3/2)/d^3-315/16 *f^2*(d*x+c)^(1/3)*sin(a+b*(d*x+c)^(2/3))/b^4/d^3+6*f*(-c*f+d*e)*(d*x+c)^( 2/3)*sin(a+b*(d*x+c)^(2/3))/b^2/d^3+21/4*f^2*(d*x+c)^(5/3)*sin(a+b*(d*x+c) ^(2/3))/b^2/d^3
Result contains complex when optimal does not.
Time = 2.92 (sec) , antiderivative size = 432, normalized size of antiderivative = 0.84 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=-\frac {3 i \left ((\cos (a)+i \sin (a)) \left ((1+i) \left (-105 i f^2+8 b^3 (d e-c f)^2\right ) \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {(1+i) \sqrt {b} \sqrt [3]{c+d x}}{\sqrt {2}}\right )+2 \sqrt {b} \left (-105 f^2 \sqrt [3]{c+d x}-8 i b^3 d^2 \sqrt [3]{c+d x} (e+f x)^2+4 b^2 f (c+d x)^{2/3} (8 d e-c f+7 d f x)+2 i b f (16 d e+19 c f+35 d f x)\right ) \left (\cos \left (b (c+d x)^{2/3}\right )+i \sin \left (b (c+d x)^{2/3}\right )\right )\right )+\left (2 \sqrt {b} \left (105 f^2 \sqrt [3]{c+d x}-8 i b^3 d^2 \sqrt [3]{c+d x} (e+f x)^2+4 b^2 f (c+d x)^{2/3} (-8 d e+c f-7 d f x)+2 i b f (16 d e+19 c f+35 d f x)\right )+(1+i) \left (105 i f^2+8 b^3 (d e-c f)^2\right ) \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {(1+i) \sqrt {b} \sqrt [3]{c+d x}}{\sqrt {2}}\right ) \left (\cos \left (b (c+d x)^{2/3}\right )+i \sin \left (b (c+d x)^{2/3}\right )\right )\right ) \left (\cos \left (a+b (c+d x)^{2/3}\right )-i \sin \left (a+b (c+d x)^{2/3}\right )\right )\right )}{64 b^{9/2} d^3} \] Input:
Integrate[(e + f*x)^2*Sin[a + b*(c + d*x)^(2/3)],x]
Output:
(((-3*I)/64)*((Cos[a] + I*Sin[a])*((1 + I)*((-105*I)*f^2 + 8*b^3*(d*e - c* f)^2)*Sqrt[Pi/2]*Erfi[((1 + I)*Sqrt[b]*(c + d*x)^(1/3))/Sqrt[2]] + 2*Sqrt[ b]*(-105*f^2*(c + d*x)^(1/3) - (8*I)*b^3*d^2*(c + d*x)^(1/3)*(e + f*x)^2 + 4*b^2*f*(c + d*x)^(2/3)*(8*d*e - c*f + 7*d*f*x) + (2*I)*b*f*(16*d*e + 19* c*f + 35*d*f*x))*(Cos[b*(c + d*x)^(2/3)] + I*Sin[b*(c + d*x)^(2/3)])) + (2 *Sqrt[b]*(105*f^2*(c + d*x)^(1/3) - (8*I)*b^3*d^2*(c + d*x)^(1/3)*(e + f*x )^2 + 4*b^2*f*(c + d*x)^(2/3)*(-8*d*e + c*f - 7*d*f*x) + (2*I)*b*f*(16*d*e + 19*c*f + 35*d*f*x)) + (1 + I)*((105*I)*f^2 + 8*b^3*(d*e - c*f)^2)*Sqrt[ Pi/2]*Erf[((1 + I)*Sqrt[b]*(c + d*x)^(1/3))/Sqrt[2]]*(Cos[b*(c + d*x)^(2/3 )] + I*Sin[b*(c + d*x)^(2/3)]))*(Cos[a + b*(c + d*x)^(2/3)] - I*Sin[a + b* (c + d*x)^(2/3)])))/(b^(9/2)*d^3)
Time = 0.66 (sec) , antiderivative size = 482, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
| \(\displaystyle \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx\) |
| \(\Big \downarrow \) 3914 |
| \(\displaystyle \frac {3 \int \left (f^2 \sin \left (a+b (c+d x)^{2/3}\right ) (c+d x)^{8/3}+2 f (d e-c f) \sin \left (a+b (c+d x)^{2/3}\right ) (c+d x)^{5/3}+(d e-c f)^2 \sin \left (a+b (c+d x)^{2/3}\right ) (c+d x)^{2/3}\right )d\sqrt [3]{c+d x}}{d^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \left (\frac {\sqrt {\frac {\pi }{2}} \cos (a) (d e-c f)^2 \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) (d e-c f)^2 \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2}}+\frac {105 \sqrt {\frac {\pi }{2}} f^2 \sin (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{16 b^{9/2}}+\frac {105 \sqrt {\frac {\pi }{2}} f^2 \cos (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [3]{c+d x}\right )}{16 b^{9/2}}-\frac {105 f^2 \sqrt [3]{c+d x} \sin \left (a+b (c+d x)^{2/3}\right )}{16 b^4}+\frac {2 f (d e-c f) \cos \left (a+b (c+d x)^{2/3}\right )}{b^3}+\frac {35 f^2 (c+d x) \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3}+\frac {2 f (c+d x)^{2/3} (d e-c f) \sin \left (a+b (c+d x)^{2/3}\right )}{b^2}+\frac {7 f^2 (c+d x)^{5/3} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2}-\frac {f (c+d x)^{4/3} (d e-c f) \cos \left (a+b (c+d x)^{2/3}\right )}{b}-\frac {\sqrt [3]{c+d x} (d e-c f)^2 \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}-\frac {f^2 (c+d x)^{7/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b}\right )}{d^3}\) |
Input:
Int[(e + f*x)^2*Sin[a + b*(c + d*x)^(2/3)],x]
Output:
(3*((2*f*(d*e - c*f)*Cos[a + b*(c + d*x)^(2/3)])/b^3 - ((d*e - c*f)^2*(c + d*x)^(1/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b) + (35*f^2*(c + d*x)*Cos[a + b*(c + d*x)^(2/3)])/(8*b^3) - (f*(d*e - c*f)*(c + d*x)^(4/3)*Cos[a + b*(c + d*x)^(2/3)])/b - (f^2*(c + d*x)^(7/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b) + ((d*e - c*f)^2*Sqrt[Pi/2]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)^( 1/3)])/(2*b^(3/2)) + (105*f^2*Sqrt[Pi/2]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi ]*(c + d*x)^(1/3)])/(16*b^(9/2)) + (105*f^2*Sqrt[Pi/2]*FresnelC[Sqrt[b]*Sq rt[2/Pi]*(c + d*x)^(1/3)]*Sin[a])/(16*b^(9/2)) - ((d*e - c*f)^2*Sqrt[Pi/2] *FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)]*Sin[a])/(2*b^(3/2)) - (105*f ^2*(c + d*x)^(1/3)*Sin[a + b*(c + d*x)^(2/3)])/(16*b^4) + (2*f*(d*e - c*f) *(c + d*x)^(2/3)*Sin[a + b*(c + d*x)^(2/3)])/b^2 + (7*f^2*(c + d*x)^(5/3)* Sin[a + b*(c + d*x)^(2/3)])/(4*b^2)))/d^3
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 = 2.93 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.71
| method | result | size |
| derivativedivides | \(\frac {-\frac {3 f^{2} \left (d x +c \right )^{\frac {7}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {21 f^{2} \left (\frac {\left (d x +c \right )^{\frac {5}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}-\frac {5 \left (-\frac {\left (d x +c \right ) \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {\frac {3 \left (d x +c \right )^{\frac {1}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{4 b}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )+\sin \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )\right )}{8 b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )}{2 b}+\frac {3 \left (c f -d e \right ) f \left (d x +c \right )^{\frac {4}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{b}-\frac {12 \left (c f -d e \right ) f \left (\frac {\left (d x +c \right )^{\frac {2}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {\cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b^{2}}\right )}{b}-\frac {3 \left (c f -d e \right )^{2} \left (d x +c \right )^{\frac {1}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {3 \left (c f -d e \right )^{2} \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )-\sin \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )\right )}{4 b^{\frac {3}{2}}}}{d^{3}}\) | \(366\) |
| default | \(\frac {-\frac {3 f^{2} \left (d x +c \right )^{\frac {7}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {21 f^{2} \left (\frac {\left (d x +c \right )^{\frac {5}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}-\frac {5 \left (-\frac {\left (d x +c \right ) \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {\frac {3 \left (d x +c \right )^{\frac {1}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{4 b}-\frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )+\sin \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )\right )}{8 b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )}{2 b}+\frac {3 \left (c f -d e \right ) f \left (d x +c \right )^{\frac {4}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{b}-\frac {12 \left (c f -d e \right ) f \left (\frac {\left (d x +c \right )^{\frac {2}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {\cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b^{2}}\right )}{b}-\frac {3 \left (c f -d e \right )^{2} \left (d x +c \right )^{\frac {1}{3}} \cos \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{2 b}+\frac {3 \left (c f -d e \right )^{2} \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )-\sin \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}\, \left (d x +c \right )^{\frac {1}{3}}}{\sqrt {\pi }}\right )\right )}{4 b^{\frac {3}{2}}}}{d^{3}}\) | \(366\) |
| parts | \(\text {Expression too large to display}\) | \(1316\) |
Input:
int((f*x+e)^2*sin(a+b*(d*x+c)^(2/3)),x,method=_RETURNVERBOSE)
Output:
3/d^3*(-1/2*f^2/b*(d*x+c)^(7/3)*cos(a+b*(d*x+c)^(2/3))+7/2*f^2/b*(1/2/b*(d *x+c)^(5/3)*sin(a+b*(d*x+c)^(2/3))-5/2/b*(-1/2/b*(d*x+c)*cos(a+b*(d*x+c)^( 2/3))+3/2/b*(1/2/b*(d*x+c)^(1/3)*sin(a+b*(d*x+c)^(2/3))-1/4/b^(3/2)*2^(1/2 )*Pi^(1/2)*(cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))+sin(a) *FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))))))+(c*f-d*e)*f/b*(d*x+c )^(4/3)*cos(a+b*(d*x+c)^(2/3))-4*(c*f-d*e)*f/b*(1/2/b*(d*x+c)^(2/3)*sin(a+ b*(d*x+c)^(2/3))+1/2/b^2*cos(a+b*(d*x+c)^(2/3)))-1/2*(c*f-d*e)^2/b*(d*x+c) ^(1/3)*cos(a+b*(d*x+c)^(2/3))+1/4*(c*f-d*e)^2/b^(3/2)*2^(1/2)*Pi^(1/2)*(co s(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))-sin(a)*FresnelS(b^(1 /2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/3))))
Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.60 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\frac {3 \, {\left (\sqrt {2} {\left (105 \, \pi f^{2} \sin \left (a\right ) + 8 \, \pi {\left (b^{3} d^{2} e^{2} - 2 \, b^{3} c d e f + b^{3} c^{2} f^{2}\right )} \cos \left (a\right )\right )} \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} {\left (d x + c\right )}^{\frac {1}{3}} \sqrt {\frac {b}{\pi }}\right ) + \sqrt {2} {\left (105 \, \pi f^{2} \cos \left (a\right ) - 8 \, \pi {\left (b^{3} d^{2} e^{2} - 2 \, b^{3} c d e f + b^{3} c^{2} f^{2}\right )} \sin \left (a\right )\right )} \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {2} {\left (d x + c\right )}^{\frac {1}{3}} \sqrt {\frac {b}{\pi }}\right ) + 4 \, {\left (35 \, b^{2} d f^{2} x + 16 \, b^{2} d e f + 19 \, b^{2} c f^{2} - 4 \, {\left (b^{4} d^{2} f^{2} x^{2} + 2 \, b^{4} d^{2} e f x + b^{4} d^{2} e^{2}\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} \cos \left ({\left (d x + c\right )}^{\frac {2}{3}} b + a\right ) - 2 \, {\left (105 \, {\left (d x + c\right )}^{\frac {1}{3}} b f^{2} - 4 \, {\left (7 \, b^{3} d f^{2} x + 8 \, b^{3} d e f - b^{3} c f^{2}\right )} {\left (d x + c\right )}^{\frac {2}{3}}\right )} \sin \left ({\left (d x + c\right )}^{\frac {2}{3}} b + a\right )\right )}}{32 \, b^{5} d^{3}} \] Input:
integrate((f*x+e)^2*sin(a+b*(d*x+c)^(2/3)),x, algorithm="fricas")
Output:
3/32*(sqrt(2)*(105*pi*f^2*sin(a) + 8*pi*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3 *c^2*f^2)*cos(a))*sqrt(b/pi)*fresnel_cos(sqrt(2)*(d*x + c)^(1/3)*sqrt(b/pi )) + sqrt(2)*(105*pi*f^2*cos(a) - 8*pi*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3* c^2*f^2)*sin(a))*sqrt(b/pi)*fresnel_sin(sqrt(2)*(d*x + c)^(1/3)*sqrt(b/pi) ) + 4*(35*b^2*d*f^2*x + 16*b^2*d*e*f + 19*b^2*c*f^2 - 4*(b^4*d^2*f^2*x^2 + 2*b^4*d^2*e*f*x + b^4*d^2*e^2)*(d*x + c)^(1/3))*cos((d*x + c)^(2/3)*b + a ) - 2*(105*(d*x + c)^(1/3)*b*f^2 - 4*(7*b^3*d*f^2*x + 8*b^3*d*e*f - b^3*c* f^2)*(d*x + c)^(2/3))*sin((d*x + c)^(2/3)*b + a))/(b^5*d^3)
\[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\int \left (e + f x\right )^{2} \sin {\left (a + b \left (c + d x\right )^{\frac {2}{3}} \right )}\, dx \] Input:
integrate((f*x+e)**2*sin(a+b*(d*x+c)**(2/3)),x)
Output:
Integral((e + f*x)**2*sin(a + b*(c + d*x)**(2/3)), x)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 561, normalized size of antiderivative = 1.09 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\text {Too large to display} \] Input:
integrate((f*x+e)^2*sin(a+b*(d*x+c)^(2/3)),x, algorithm="maxima")
Output:
-3/128*(8*(sqrt(2)*sqrt(pi)*(((I - 1)*cos(a) + (I + 1)*sin(a))*erf((d*x + c)^(1/3)*sqrt(I*b)) + (-(I + 1)*cos(a) - (I - 1)*sin(a))*erf((d*x + c)^(1/ 3)*sqrt(-I*b)))*b^(3/2) + 8*(d*x + c)^(1/3)*b^2*cos((d*x + c)^(2/3)*b + a) )*e^2/b^3 - 16*(sqrt(2)*sqrt(pi)*(((I - 1)*cos(a) + (I + 1)*sin(a))*erf((d *x + c)^(1/3)*sqrt(I*b)) + (-(I + 1)*cos(a) - (I - 1)*sin(a))*erf((d*x + c )^(1/3)*sqrt(-I*b)))*b^(3/2) + 8*(d*x + c)^(1/3)*b^2*cos((d*x + c)^(2/3)*b + a))*c*e*f/(b^3*d) + 8*(sqrt(2)*sqrt(pi)*(((I - 1)*cos(a) + (I + 1)*sin( a))*erf((d*x + c)^(1/3)*sqrt(I*b)) + (-(I + 1)*cos(a) - (I - 1)*sin(a))*er f((d*x + c)^(1/3)*sqrt(-I*b)))*b^(3/2) + 8*(d*x + c)^(1/3)*b^2*cos((d*x + c)^(2/3)*b + a))*c^2*f^2/(b^3*d^2) - 128*(2*(d*x + c)^(2/3)*b*sin((d*x + c )^(2/3)*b + a) - ((d*x + c)^(4/3)*b^2 - 2)*cos((d*x + c)^(2/3)*b + a))*e*f /(b^3*d) + 128*(2*(d*x + c)^(2/3)*b*sin((d*x + c)^(2/3)*b + a) - ((d*x + c )^(4/3)*b^2 - 2)*cos((d*x + c)^(2/3)*b + a))*c*f^2/(b^3*d^2) - (105*sqrt(2 )*sqrt(pi)*(((I + 1)*cos(a) - (I - 1)*sin(a))*erf((d*x + c)^(1/3)*sqrt(I*b )) + (-(I - 1)*cos(a) + (I + 1)*sin(a))*erf((d*x + c)^(1/3)*sqrt(-I*b)))*b ^(3/2) - 16*(4*(d*x + c)^(7/3)*b^5 - 35*(d*x + c)*b^3)*cos((d*x + c)^(2/3) *b + a) + 56*(4*(d*x + c)^(5/3)*b^4 - 15*(d*x + c)^(1/3)*b^2)*sin((d*x + c )^(2/3)*b + a))*f^2/(b^6*d^2))/d
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 770, normalized size of antiderivative = 1.50 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\text {Too large to display} \] Input:
integrate((f*x+e)^2*sin(a+b*(d*x+c)^(2/3)),x, algorithm="giac")
Output:
-3/64*(8*e^2*(sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(-I*b/abs( b) + 1)*sqrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(b) + 1)*sqrt(abs(b))) + sqrt(2) *sqrt(pi)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(I*b/abs(b) + 1)*sqrt(abs(b)))* e^(-I*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b))) + 2*(d*x + c)^(1/3)*e^(I*(d*x + c)^(2/3)*b + I*a)/b + 2*(d*x + c)^(1/3)*e^(-I*(d*x + c)^(2/3)*b - I*a)/b) - 16*(sqrt(2)*sqrt(pi)*c*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(-I*b/abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(b) + 1)*sqrt(abs(b))) + sqrt(2)*sqrt (pi)*c*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^( -I*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b))) + 2*I*(I*(d*x + c)^(4/3)*b^2 - I*( d*x + c)^(1/3)*b^2*c - 2*(d*x + c)^(2/3)*b - 2*I)*e^(I*(d*x + c)^(2/3)*b + I*a)/b^3 + 2*I*(I*(d*x + c)^(4/3)*b^2 - I*(d*x + c)^(1/3)*b^2*c + 2*(d*x + c)^(2/3)*b - 2*I)*e^(-I*(d*x + c)^(2/3)*b - I*a)/b^3)*e*f/d + f^2*(I*sqr t(2)*sqrt(pi)*(-8*I*b^3*c^2 - 105)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(-I*b/ abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/(b^4*(-I*b/abs(b) + 1)*sqrt(abs(b))) + I *sqrt(2)*sqrt(pi)*(-8*I*b^3*c^2 + 105)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(I *b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/(b^4*(I*b/abs(b) + 1)*sqrt(abs(b))) - 2*I*(8*I*(d*x + c)^(7/3)*b^3 - 16*I*(d*x + c)^(4/3)*b^3*c + 8*I*(d*x + c )^(1/3)*b^3*c^2 - 28*(d*x + c)^(5/3)*b^2 + 32*(d*x + c)^(2/3)*b^2*c + 70*( -I*d*x - I*c)*b + 32*I*b*c + 105*(d*x + c)^(1/3))*e^(I*(d*x + c)^(2/3)*b + I*a)/b^4 - 2*I*(8*I*(d*x + c)^(7/3)*b^3 - 16*I*(d*x + c)^(4/3)*b^3*c +...
Timed out. \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^{2/3}\right )\,{\left (e+f\,x\right )}^2 \,d x \] Input:
int(sin(a + b*(c + d*x)^(2/3))*(e + f*x)^2,x)
Output:
int(sin(a + b*(c + d*x)^(2/3))*(e + f*x)^2, x)
\[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{2/3}\right ) \, dx=\left (\int \sin \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right )d x \right ) e^{2}+\left (\int \sin \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right ) x^{2}d x \right ) f^{2}+2 \left (\int \sin \left (\left (d x +c \right )^{\frac {2}{3}} b +a \right ) x d x \right ) e f \] Input:
int((f*x+e)^2*sin(a+b*(d*x+c)^(2/3)),x)
Output:
int(sin((c + d*x)**(2/3)*b + a),x)*e**2 + int(sin((c + d*x)**(2/3)*b + a)* x**2,x)*f**2 + 2*int(sin((c + d*x)**(2/3)*b + a)*x,x)*e*f