Integrand size = 27, antiderivative size = 262 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {2 b \sqrt [3]{c+d x} (e (c+d x))^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d}+\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{5 d (c+d x)^{2/3}}-\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{5 d (c+d x)^{2/3}}-\frac {4 b^2 (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d \sqrt [3]{c+d x}}+\frac {3 (c+d x) (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d} \]
2/5*b*(d*x+c)^(1/3)*(e*(d*x+c))^(2/3)*cos(a+b/(d*x+c)^(2/3))/d-4/5*b^2*(e* (d*x+c))^(2/3)*sin(a+b/(d*x+c)^(2/3))/d/(d*x+c)^(1/3)+3/5*(d*x+c)*(e*(d*x+ c))^(2/3)*sin(a+b/(d*x+c)^(2/3))/d+4/5*b^(5/2)*(e*(d*x+c))^(2/3)*cos(a)*Fr esnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*2^(1/2)*Pi^(1/2)/d/(d*x+c)^ (2/3)-4/5*b^(5/2)*(e*(d*x+c))^(2/3)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x +c)^(1/3))*sin(a)*2^(1/2)*Pi^(1/2)/d/(d*x+c)^(2/3)
Time = 0.31 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.87 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {(e (c+d x))^{2/3} \left (2 b c \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+2 b d x \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+4 b^{5/2} \sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-4 b^{5/2} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)-4 b^2 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+3 c (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+3 d x (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{5 d (c+d x)^{2/3}} \]
((e*(c + d*x))^(2/3)*(2*b*c*Cos[a + b/(c + d*x)^(2/3)] + 2*b*d*x*Cos[a + b /(c + d*x)^(2/3)] + 4*b^(5/2)*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/P i])/(c + d*x)^(1/3)] - 4*b^(5/2)*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/ (c + d*x)^(1/3)]*Sin[a] - 4*b^2*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(2/3)] + 3*c*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)] + 3*d*x*(c + d*x)^(2/3)* Sin[a + b/(c + d*x)^(2/3)]))/(5*d*(c + d*x)^(2/3))
Time = 0.68 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.78, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3916, 3898, 3896, 3890, 3868, 3869, 3868, 3835, 3832, 3833}
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.
\(\displaystyle \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx\) |
\(\Big \downarrow \) 3916 |
\(\displaystyle \frac {\int (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d(c+d x)}{d}\) |
\(\Big \downarrow \) 3898 |
\(\displaystyle \frac {(e (c+d x))^{2/3} \int (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d(c+d x)}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3896 |
\(\displaystyle \frac {3 (e (c+d x))^{2/3} \int (c+d x)^{4/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\sqrt [3]{c+d x}}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3890 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \int \frac {\sin \left (a+b (c+d x)^{2/3}\right )}{(c+d x)^2}d\frac {1}{\sqrt [3]{c+d x}}}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \int \frac {\cos \left (a+b (c+d x)^{2/3}\right )}{(c+d x)^{4/3}}d\frac {1}{\sqrt [3]{c+d x}}-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3869 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \left (-\frac {2}{3} b \int \frac {\sin \left (a+b (c+d x)^{2/3}\right )}{(c+d x)^{2/3}}d\frac {1}{\sqrt [3]{c+d x}}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \left (-\frac {2}{3} b \left (2 b \int \cos \left (a+b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3835 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \left (-\frac {2}{3} b \left (2 b \left (\cos (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\sin (a) \int \sin \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \left (-\frac {2}{3} b \left (2 b \left (\cos (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle -\frac {3 (e (c+d x))^{2/3} \left (\frac {2}{5} b \left (-\frac {2}{3} b \left (2 b \left (\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{5 (c+d x)^{5/3}}\right )}{d (c+d x)^{2/3}}\) |
(-3*(e*(c + d*x))^(2/3)*(-1/5*Sin[a + b*(c + d*x)^(2/3)]/(c + d*x)^(5/3) + (2*b*(-1/3*Cos[a + b*(c + d*x)^(2/3)]/(c + d*x) - (2*b*(2*b*((Sqrt[Pi/2]* Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/Sqrt[b] - (Sqrt[Pi/ 2]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/Sqrt[b]) - Sin[a + b*(c + d*x)^(2/3)]/(c + d*x)^(1/3)))/3))/5))/(d*(c + d*x)^(2/3))
3.3.50.3.1 Defintions of rubi rules used
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[Cos[c] In t[Cos[d*(e + f*x)^2], x], x] - Simp[Sin[c] Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]
Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x) ^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1))), x] - Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[m, -1]
Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x) ^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1))), x] + Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[m, -1]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> -Subst[Int[(a + b*Sin[c + d/x^n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a , b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Module[{k = Denominator[n]}, Simp[k Subst[Int[x^(k*(m + 1) - 1)*(a + b*Sin[c + d*x^(k*n)])^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[p] && FractionQ[n]
Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_ Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IntegerQ[ p] && FractionQ[n]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f _.)*(x_))^(n_)])^(p_.), x_Symbol] :> Simp[1/f Subst[Int[(h*(x/f))^m*(a + b*Sin[c + d*x^n])^p, x], x, e + f*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && EqQ[f*g - e*h, 0]
\[\int \left (d e x +c e \right )^{\frac {2}{3}} \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )d x\]
\[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int { {\left (d e x + c e\right )}^{\frac {2}{3}} \sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) \,d x } \]
\[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int \left (e \left (c + d x\right )\right )^{\frac {2}{3}} \sin {\left (a + \frac {b}{\left (c + d x\right )^{\frac {2}{3}}} \right )}\, dx \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.77 (sec) , antiderivative size = 749, normalized size of antiderivative = 2.86 \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\text {Too large to display} \]
-3/8*((((-I*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-5/2, - I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (I*gamma(-5/ 2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(-5/2, I*b/(d*x + c)^(2/3))) *cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan 2(0, d*x + c)) - (gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(-5 /2, I*b/(d*x + c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*cos(a) - ((gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(-5/2, -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (gamma(-5/2, -I*b*conju gate((d*x + c)^(-2/3))) + gamma(-5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) - (-I*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3)) ) + I*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + c)) - (-I*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-5/2, I *b/(d*x + c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*sin(a))*d*e^( 2/3)*x + (((-I*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-5/2 , -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (I*gamma( -5/2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(-5/2, I*b/(d*x + c)^(2/3 )))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (gamma(-5/2, I*b*conjugate((d *x + c)^(-2/3))) + gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arc tan2(0, d*x + c)) - (gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + ga...
\[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int { {\left (d e x + c e\right )}^{\frac {2}{3}} \sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) \,d x } \]
Timed out. \[ \int (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )\,{\left (c\,e+d\,e\,x\right )}^{2/3} \,d x \]