Integrand size = 27, antiderivative size = 141 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=-\frac {3 \sqrt {\pi } \sqrt [3]{c+d x} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {2} \sqrt {b} d e \sqrt [3]{e (c+d x)}}-\frac {3 \sqrt {\pi } \sqrt [3]{c+d x} \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{\sqrt {2} \sqrt {b} d e \sqrt [3]{e (c+d x)}} \]
-3/2*(d*x+c)^(1/3)*cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3)) *Pi^(1/2)/d/e/(e*(d*x+c))^(1/3)*2^(1/2)/b^(1/2)-3/2*(d*x+c)^(1/3)*FresnelC (b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(a)*Pi^(1/2)/d/e/(e*(d*x+c))^( 1/3)*2^(1/2)/b^(1/2)
Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.68 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=-\frac {3 \sqrt {\frac {\pi }{2}} (c+d x)^{4/3} \left (\cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+\operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)\right )}{\sqrt {b} d (e (c+d x))^{4/3}} \]
(-3*Sqrt[Pi/2]*(c + d*x)^(4/3)*(Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] + FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a]))/(Sqr t[b]*d*(e*(c + d*x))^(4/3))
Time = 0.42 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3916, 3898, 3864, 3834, 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.
\(\displaystyle \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx\) |
\(\Big \downarrow \) 3916 |
\(\displaystyle \frac {\int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(e (c+d x))^{4/3}}d(c+d x)}{d}\) |
\(\Big \downarrow \) 3898 |
\(\displaystyle \frac {\sqrt [3]{c+d x} \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c+d x)^{4/3}}d(c+d x)}{d e \sqrt [3]{e (c+d x)}}\) |
\(\Big \downarrow \) 3864 |
\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{\sqrt [3]{c+d x}}}{d e \sqrt [3]{e (c+d x)}}\) |
\(\Big \downarrow \) 3834 |
\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\sin (a) \int \cos \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{\sqrt [3]{c+d x}}+\cos (a) \int \sin \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{\sqrt [3]{c+d x}}\right )}{d e \sqrt [3]{e (c+d x)}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\sin (a) \int \cos \left (\frac {b}{(c+d x)^{2/3}}\right )d\frac {1}{\sqrt [3]{c+d x}}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )}{d e \sqrt [3]{e (c+d x)}}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle -\frac {3 \sqrt [3]{c+d x} \left (\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )}{d e \sqrt [3]{e (c+d x)}}\) |
(-3*(c + d*x)^(1/3)*((Sqrt[Pi/2]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/Sqrt[b] + (Sqrt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x )^(1/3)]*Sin[a])/Sqrt[b]))/(d*e*(e*(c + d*x))^(1/3))
3.3.54.3.1 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[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[Sin[c] In t[Cos[d*(e + f*x)^2], x], x] + Simp[Cos[c] Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]
Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)], x_Symbol] :> Simp[2/n Subst[ Int[Sin[a + b*x^2], x], x, x^(n/2)], x] /; FreeQ[{a, b, m, n}, x] && EqQ[m, n/2 - 1]
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]
\[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {4}{3}}} \,d x } \]
integral((d*e*x + c*e)^(2/3)*sin((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)
\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=\int \frac {\sin {\left (a + \frac {b}{\left (c + d x\right )^{\frac {2}{3}}} \right )}}{\left (e \left (c + d x\right )\right )^{\frac {4}{3}}}\, dx \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.50 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.45 \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=\text {Too large to display} \]
3/8*(((-I*sqrt(pi)*(erf(sqrt(I*b*conjugate((d*x + c)^(-2/3)))) - 1) + I*sq rt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(1/4*pi + 1/3*arctan2(0, d*x + c)) + (I*sqrt(pi)*(erf(sqrt(-I*b*conjugate((d*x + c)^(-2/3)))) - 1) - I*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1))*cos(-1/4*pi + 1/3*arcta n2(0, d*x + c)) - (sqrt(pi)*(erf(sqrt(I*b*conjugate((d*x + c)^(-2/3)))) - 1) + sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*sin(1/4*pi + 1/3*arct an2(0, d*x + c)) + (sqrt(pi)*(erf(sqrt(-I*b*conjugate((d*x + c)^(-2/3)))) - 1) + sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1))*sin(-1/4*pi + 1/3*ar ctan2(0, d*x + c)))*cos(a) - ((sqrt(pi)*(erf(sqrt(I*b*conjugate((d*x + c)^ (-2/3)))) - 1) + sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(1/4*p i + 1/3*arctan2(0, d*x + c)) + (sqrt(pi)*(erf(sqrt(-I*b*conjugate((d*x + c )^(-2/3)))) - 1) + sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1))*cos(-1/4 *pi + 1/3*arctan2(0, d*x + c)) - (I*sqrt(pi)*(erf(sqrt(I*b*conjugate((d*x + c)^(-2/3)))) - 1) - I*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*si n(1/4*pi + 1/3*arctan2(0, d*x + c)) - (I*sqrt(pi)*(erf(sqrt(-I*b*conjugate ((d*x + c)^(-2/3)))) - 1) - I*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1 ))*sin(-1/4*pi + 1/3*arctan2(0, d*x + c)))*sin(a))/((d*x + c)^(1/3)*d*e^(4 /3)*sqrt(b/(d*x + c)^(2/3)))
\[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=\int { \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{4/3}} \, dx=\int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{4/3}} \,d x \]