3.3.53 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{2/3}} \, dx\) [253]

Optimal. Leaf size=164 \[ -\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3516, 3498, 3496, 3440, 3468, 3435, 3433, 3432} \begin {gather*} -\frac {3 \sqrt {2 \pi } \sqrt {b} \cos (a) (c+d x)^{2/3} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 \sqrt {2 \pi } \sqrt {b} \sin (a) (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3432

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 3433

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 3435

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3440

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(
a + b*Sin[c + d/x^n])^p/x^2, x], x, 1/(e + f*x)], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[n,
0] && EqQ[n, -2]

Rule 3468

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] - Dist[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 3496

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Module[{k = Denominator[n]}, D
ist[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]

Rule 3498

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[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] && Integ
erQ[p] && FractionQ[n]

Rule 3516

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[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]

Rubi steps

\begin {align*} \int \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{2/3}} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{(e x)^{2/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x)^{2/3} \text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{x^{2/3}} \, dx,x,c+d x\right )}{d (e (c+d x))^{2/3}}\\ &=\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \sin \left (a+\frac {b}{x^2}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d (e (c+d x))^{2/3}}\\ &=-\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \frac {\sin \left (a+b x^2\right )}{x^2} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}-\frac {\left (6 b (c+d x)^{2/3}\right ) \text {Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}-\frac {\left (6 b (c+d x)^{2/3} \cos (a)\right ) \text {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {\left (6 b (c+d x)^{2/3} \sin (a)\right ) \text {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}\\ &=-\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d (e (c+d x))^{2/3}}+\frac {3 \sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d (e (c+d x))^{2/3}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d (e (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 136, normalized size = 0.83 \begin {gather*} \frac {3 \left (-\sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+\sqrt {b} \sqrt {2 \pi } (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)+(c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{d (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.57, size = 382, normalized size = 2.33 \begin {gather*} -\frac {3 \, {\left (d x + c\right )}^{\frac {1}{3}} {\left ({\left ({\left (-i \, \Gamma \left (-\frac {1}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (-\frac {1}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (i \, \Gamma \left (-\frac {1}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - i \, \Gamma \left (-\frac {1}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (\Gamma \left (-\frac {1}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-\frac {1}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (\Gamma \left (-\frac {1}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-\frac {1}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \cos \left (a\right ) - {\left ({\left (\Gamma \left (-\frac {1}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-\frac {1}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (\Gamma \left (-\frac {1}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-\frac {1}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (-i \, \Gamma \left (-\frac {1}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (-\frac {1}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (-i \, \Gamma \left (-\frac {1}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (-\frac {1}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \sin \left (a\right )\right )} \sqrt {\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}} e^{\left (-\frac {2}{3}\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*(d*x + c)^(1/3)*(((-I*gamma(-1/2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-1/2, -I*b/(d*x + c)^(2/3)))
*cos(1/4*pi + 1/3*arctan2(0, d*x + c)) + (I*gamma(-1/2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(-1/2, I*b/
(d*x + c)^(2/3)))*cos(-1/4*pi + 1/3*arctan2(0, d*x + c)) + (gamma(-1/2, I*b*conjugate((d*x + c)^(-2/3))) + gam
ma(-1/2, -I*b/(d*x + c)^(2/3)))*sin(1/4*pi + 1/3*arctan2(0, d*x + c)) - (gamma(-1/2, -I*b*conjugate((d*x + c)^
(-2/3))) + gamma(-1/2, I*b/(d*x + c)^(2/3)))*sin(-1/4*pi + 1/3*arctan2(0, d*x + c)))*cos(a) - ((gamma(-1/2, I*
b*conjugate((d*x + c)^(-2/3))) + gamma(-1/2, -I*b/(d*x + c)^(2/3)))*cos(1/4*pi + 1/3*arctan2(0, d*x + c)) + (g
amma(-1/2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1/2, I*b/(d*x + c)^(2/3)))*cos(-1/4*pi + 1/3*arctan2(0,
d*x + c)) - (-I*gamma(-1/2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-1/2, -I*b/(d*x + c)^(2/3)))*sin(1/4*pi
+ 1/3*arctan2(0, d*x + c)) - (-I*gamma(-1/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-1/2, I*b/(d*x + c)^
(2/3)))*sin(-1/4*pi + 1/3*arctan2(0, d*x + c)))*sin(a))*sqrt(b/(d*x + c)^(2/3))*e^(-2/3)/d

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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