\(\int \sin (a+\frac {b}{(c+d x)^{2/3}}) \, dx\) [224]

Optimal result

Integrand size = 14, antiderivative size = 141 \[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {2 b \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d}+\frac {2 b^{3/2} \sqrt {2 \pi } \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d}+\frac {2 b^{3/2} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d}+\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d} \] Output:

2*b*(d*x+c)^(1/3)*cos(a+b/(d*x+c)^(2/3))/d+2*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))/d+2*b^(3/2)*2^(1/2)*Pi 
^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(a)/d+(d*x+c)*s 
in(a+b/(d*x+c)^(2/3))/d
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04 \[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {2 b \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+2 b^{3/2} \sqrt {2 \pi } \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+2 b^{3/2} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)+c \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+d x \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d} \] Input:

Integrate[Sin[a + b/(c + d*x)^(2/3)],x]
 

Output:

(2*b*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)] + 2*b^(3/2)*Sqrt[2*Pi]*Cos 
[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] + 2*b^(3/2)*Sqrt[2*Pi]* 
FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] + c*Sin[a + b/(c + d 
*x)^(2/3)] + d*x*Sin[a + b/(c + d*x)^(2/3)])/d
 

Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3844, 3890, 3868, 3869, 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.

\(\displaystyle \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx\)

\(\Big \downarrow \) 3844

\(\displaystyle \frac {3 \int (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 3890

\(\displaystyle -\frac {3 \int \frac {\sin \left (a+b (c+d x)^{2/3}\right )}{(c+d x)^{4/3}}d\frac {1}{\sqrt [3]{c+d x}}}{d}\)

\(\Big \downarrow \) 3868

\(\displaystyle -\frac {3 \left (\frac {2}{3} b \int \frac {\cos \left (a+b (c+d x)^{2/3}\right )}{(c+d x)^{2/3}}d\frac {1}{\sqrt [3]{c+d x}}-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 3869

\(\displaystyle -\frac {3 \left (\frac {2}{3} b \left (-2 b \int \sin \left (a+b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 3834

\(\displaystyle -\frac {3 \left (\frac {2}{3} b \left (-2 b \left (\sin (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}+\cos (a) \int \sin \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 3832

\(\displaystyle -\frac {3 \left (\frac {2}{3} b \left (-2 b \left (\sin (a) \int \cos \left (b (c+d x)^{2/3}\right )d\frac {1}{\sqrt [3]{c+d x}}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )}{d}\)

\(\Big \downarrow \) 3833

\(\displaystyle -\frac {3 \left (\frac {2}{3} b \left (-2 b \left (\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{\sqrt {b}}\right )-\frac {\cos \left (a+b (c+d x)^{2/3}\right )}{\sqrt [3]{c+d x}}\right )-\frac {\sin \left (a+b (c+d x)^{2/3}\right )}{3 (c+d x)}\right )}{d}\)

Input:

Int[Sin[a + b/(c + d*x)^(2/3)],x]
 

Output:

(-3*((2*b*(-(Cos[a + b*(c + d*x)^(2/3)]/(c + d*x)^(1/3)) - 2*b*((Sqrt[Pi/2 
]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/Sqrt[b] + (Sqrt[P 
i/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/Sqrt[b])))/3 - 
 Sin[a + b*(c + d*x)^(2/3)]/(3*(c + d*x))))/d
 

Defintions of rubi rules used

rule 3832

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]
 

rule 3833

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]
 

rule 3834

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]
 

rule 3844

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_S 
ymbol] :> Module[{k = Denominator[n]}, Simp[k/f   Subst[Int[x^(k - 1)*(a + 
b*Sin[c + d*x^(k*n)])^p, x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, 
 e, f}, x] && IGtQ[p, 0] && FractionQ[n]
 

rule 3868

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]
 

rule 3869

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]
 

rule 3890

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]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-2 b \left (-\left (d x +c \right )^{\frac {1}{3}} \cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )+\sin \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )\right )\right )}{d}\)	\(105\)
default	\(\frac {\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-2 b \left (-\left (d x +c \right )^{\frac {1}{3}} \cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )+\sin \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )\right )\right )}{d}\)	\(105\)

Input:

int(sin(a+b/(d*x+c)^(2/3)),x,method=_RETURNVERBOSE)
 

Output:

3/d*(1/3*(d*x+c)*sin(a+b/(d*x+c)^(2/3))-2/3*b*(-(d*x+c)^(1/3)*cos(a+b/(d*x 
+c)^(2/3))-b^(1/2)*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)) 
)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01 \[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {2 \, \sqrt {2} \pi b \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + 2 \, \sqrt {2} \pi b \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) \sin \left (a\right ) + 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right ) + {\left (d x + c\right )} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{d} \] Input:

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="fricas")
 

Output:

(2*sqrt(2)*pi*b*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*sqrt(b/pi)/(d*x + c) 
^(1/3)) + 2*sqrt(2)*pi*b*sqrt(b/pi)*fresnel_cos(sqrt(2)*sqrt(b/pi)/(d*x + 
c)^(1/3))*sin(a) + 2*(d*x + c)^(1/3)*b*cos((a*d*x + a*c + (d*x + c)^(1/3)* 
b)/(d*x + c)) + (d*x + c)*sin((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c)) 
)/d
 

Sympy [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int \sin {\left (a + \frac {b}{\left (c + d x\right )^{\frac {2}{3}}} \right )}\, dx \] Input:

integrate(sin(a+b/(d*x+c)**(2/3)),x)
 

Output:

Integral(sin(a + b/(c + d*x)**(2/3)), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.55 \[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\frac {\sqrt {2} {\left (2 \, \sqrt {2} {\left (d x + c\right )}^{\frac {2}{3}} \sqrt {\frac {1}{{\left (d x + c\right )}^{\frac {4}{3}}}} b^{2} \cos \left (\frac {{\left (d x + c\right )}^{\frac {2}{3}} a + b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \sqrt {2} {\left (d x + c\right )}^{\frac {4}{3}} \sqrt {\frac {1}{{\left (d x + c\right )}^{\frac {4}{3}}}} b \sin \left (\frac {{\left (d x + c\right )}^{\frac {2}{3}} a + b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + {\left ({\left (\left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )}\right )} \cos \left (a\right ) + {\left (-\left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )} + \left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )}\right )} \sin \left (a\right )\right )} b^{2} \left (\frac {b^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}}\right )^{\frac {1}{4}}\right )} \sqrt {{\left (d x + c\right )}^{\frac {4}{3}}}}{2 \, {\left (d x + c\right )}^{\frac {1}{3}} b d} \] Input:

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="maxima")
 

Output:

1/2*sqrt(2)*(2*sqrt(2)*(d*x + c)^(2/3)*sqrt((d*x + c)^(-4/3))*b^2*cos(((d* 
x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + sqrt(2)*(d*x + c)^(4/3)*sqrt((d*x + 
 c)^(-4/3))*b*sin(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + (((I + 1)*sqr 
t(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - (I - 1)*sqrt(pi)*(erf(sqrt(-I 
*b/(d*x + c)^(2/3))) - 1))*cos(a) + (-(I - 1)*sqrt(pi)*(erf(sqrt(I*b/(d*x 
+ c)^(2/3))) - 1) + (I + 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1) 
)*sin(a))*b^2*(b^2/(d*x + c)^(4/3))^(1/4))*sqrt((d*x + c)^(4/3))/((d*x + c 
)^(1/3)*b*d)
 

Giac [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int { \sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) \,d x } \] Input:

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="giac")
 

Output:

integrate(sin(a + b/(d*x + c)^(2/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right ) \,d x \] Input:

int(sin(a + b/(c + d*x)^(2/3)),x)
 

Output:

int(sin(a + b/(c + d*x)^(2/3)), x)
 

Reduce [F]

\[ \int \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx=\int \sin \left (\frac {\left (d x +c \right )^{\frac {2}{3}} a +b}{\left (d x +c \right )^{\frac {2}{3}}}\right )d x \] Input:

int(sin(a+b/(d*x+c)^(2/3)),x)
 

Output:

int(sin(((c + d*x)**(2/3)*a + b)/(c + d*x)**(2/3)),x)