3.235 \(\int (c e+d e x)^{4/3} \sin (a+b (c+d x)^{2/3}) \, dx\)

Optimal. Leaf size=267 \[ -\frac{45 \sqrt{\pi } e \cos (a) \sqrt [3]{e (c+d x)} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{45 \sqrt{\pi } e \sin (a) \sqrt [3]{e (c+d x)} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}+\frac{45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d} \]

[Out]

(45*e*(e*(c + d*x))^(1/3)*Cos[a + b*(c + d*x)^(2/3)])/(8*b^3*d) - (3*e*(c + d*x)^(4/3)*(e*(c + d*x))^(1/3)*Cos
[a + b*(c + d*x)^(2/3)])/(2*b*d) - (45*e*Sqrt[Pi]*(e*(c + d*x))^(1/3)*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c +
d*x)^(1/3)])/(8*Sqrt[2]*b^(7/2)*d*(c + d*x)^(1/3)) + (45*e*Sqrt[Pi]*(e*(c + d*x))^(1/3)*FresnelS[Sqrt[b]*Sqrt[
2/Pi]*(c + d*x)^(1/3)]*Sin[a])/(8*Sqrt[2]*b^(7/2)*d*(c + d*x)^(1/3)) + (15*e*(c + d*x)^(2/3)*(e*(c + d*x))^(1/
3)*Sin[a + b*(c + d*x)^(2/3)])/(4*b^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.273075, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {3435, 3417, 3415, 3385, 3386, 3354, 3352, 3351} \[ -\frac{45 \sqrt{\pi } e \cos (a) \sqrt [3]{e (c+d x)} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{45 \sqrt{\pi } e \sin (a) \sqrt [3]{e (c+d x)} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}+\frac{45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(4/3)*Sin[a + b*(c + d*x)^(2/3)],x]

[Out]

(45*e*(e*(c + d*x))^(1/3)*Cos[a + b*(c + d*x)^(2/3)])/(8*b^3*d) - (3*e*(c + d*x)^(4/3)*(e*(c + d*x))^(1/3)*Cos
[a + b*(c + d*x)^(2/3)])/(2*b*d) - (45*e*Sqrt[Pi]*(e*(c + d*x))^(1/3)*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c +
d*x)^(1/3)])/(8*Sqrt[2]*b^(7/2)*d*(c + d*x)^(1/3)) + (45*e*Sqrt[Pi]*(e*(c + d*x))^(1/3)*FresnelS[Sqrt[b]*Sqrt[
2/Pi]*(c + d*x)^(1/3)]*Sin[a])/(8*Sqrt[2]*b^(7/2)*d*(c + d*x)^(1/3)) + (15*e*(c + d*x)^(2/3)*(e*(c + d*x))^(1/
3)*Sin[a + b*(c + d*x)^(2/3)])/(4*b^2*d)

Rule 3435

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]

Rule 3417

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 3415

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 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3354

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 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int (c e+d e x)^{4/3} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^{4/3} \sin \left (a+b x^{2/3}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (e \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^{4/3} \sin \left (a+b x^{2/3}\right ) \, dx,x,c+d x\right )}{d \sqrt [3]{c+d x}}\\ &=\frac{\left (3 e \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^6 \sin \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d \sqrt [3]{c+d x}}\\ &=-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac{\left (15 e \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^4 \cos \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{2 b d \sqrt [3]{c+d x}}\\ &=-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}-\frac{\left (45 e \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int x^2 \sin \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{4 b^2 d \sqrt [3]{c+d x}}\\ &=\frac{45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}-\frac{\left (45 e \sqrt [3]{e (c+d x)}\right ) \operatorname{Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{8 b^3 d \sqrt [3]{c+d x}}\\ &=\frac{45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}-\frac{\left (45 e \sqrt [3]{e (c+d x)} \cos (a)\right ) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{8 b^3 d \sqrt [3]{c+d x}}+\frac{\left (45 e \sqrt [3]{e (c+d x)} \sin (a)\right ) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{8 b^3 d \sqrt [3]{c+d x}}\\ &=\frac{45 e \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{8 b^3 d}-\frac{3 e (c+d x)^{4/3} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}-\frac{45 e \sqrt{\pi } \sqrt [3]{e (c+d x)} \cos (a) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{45 e \sqrt{\pi } \sqrt [3]{e (c+d x)} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{8 \sqrt{2} b^{7/2} d \sqrt [3]{c+d x}}+\frac{15 e (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{4 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.738247, size = 175, normalized size = 0.66 \[ -\frac{3 (e (c+d x))^{4/3} \left (2 \sqrt{b} \left (\sqrt [3]{c+d x} \left (4 b^2 (c+d x)^{4/3}-15\right ) \cos \left (a+b (c+d x)^{2/3}\right )-10 b (c+d x) \sin \left (a+b (c+d x)^{2/3}\right )\right )+15 \sqrt{2 \pi } \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )-15 \sqrt{2 \pi } \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )\right )}{16 b^{7/2} d (c+d x)^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(4/3)*Sin[a + b*(c + d*x)^(2/3)],x]

[Out]

(-3*(e*(c + d*x))^(4/3)*(15*Sqrt[2*Pi]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)] - 15*Sqrt[2*Pi]*Fre
snelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)]*Sin[a] + 2*Sqrt[b]*((c + d*x)^(1/3)*(-15 + 4*b^2*(c + d*x)^(4/3))*Co
s[a + b*(c + d*x)^(2/3)] - 10*b*(c + d*x)*Sin[a + b*(c + d*x)^(2/3)])))/(16*b^(7/2)*d*(c + d*x)^(4/3))

________________________________________________________________________________________

Maple [F]  time = 0.032, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{{\frac{4}{3}}}\sin \left ( a+b \left ( dx+c \right ) ^{{\frac{2}{3}}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x)

[Out]

int((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="maxima")

[Out]

Exception raised: IndexError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d e x + c e\right )}^{\frac{4}{3}} \sin \left ({\left (d x + c\right )}^{\frac{2}{3}} b + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="fricas")

[Out]

integral((d*e*x + c*e)^(4/3)*sin((d*x + c)^(2/3)*b + a), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(4/3)*sin(a+b*(d*x+c)**(2/3)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [C]  time = 1.33883, size = 639, normalized size = 2.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(4/3)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="giac")

[Out]

-3/32*((-8*I*(-I*(d*x*e + c*e)^(2/3)*b*c*e^(-2/3) + c)*e^(I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) + I*a + 7/3)/b^2 -
8*I*(-I*(d*x*e + c*e)^(2/3)*b*c*e^(-2/3) - c)*e^(-I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) - I*a + 7/3)/b^2 - 2*I*(4*I
*(d*x*e + c*e)^(5/3)*b^2*e^(-4/3) - 10*(d*x*e + c*e)*b*e^(-2/3) - 15*I*(d*x*e + c*e)^(1/3))*e^(I*(d*x*e + c*e)
^(2/3)*b*e^(-2/3) + I*a + 2)/b^3 - 2*I*(4*I*(d*x*e + c*e)^(5/3)*b^2*e^(-4/3) + 10*(d*x*e + c*e)*b*e^(-2/3) - 1
5*I*(d*x*e + c*e)^(1/3))*e^(-I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) - I*a + 2)/b^3 - 15*sqrt(pi)*erf(-(d*x*e + c*e)^
(1/3)*sqrt(-I*b*e^(-2/3)))*e^(I*a + 2)/(sqrt(-I*b*e^(-2/3))*b^3) - 15*sqrt(pi)*erf(-(d*x*e + c*e)^(1/3)*sqrt(I
*b*e^(-2/3)))*e^(-I*a + 2)/(sqrt(I*b*e^(-2/3))*b^3))*e^(-1) + 8*((d*x*e + c*e)^(2/3)*b*cos((d*x*e + c*e)^(2/3)
*b*e^(-2/3) + a)*e^(-2/3) + (d*x*e + c*e)^(2/3)*b*cos(-(d*x*e + c*e)^(2/3)*b*e^(-2/3) - a)*e^(-2/3) - sin((d*x
*e + c*e)^(2/3)*b*e^(-2/3) + a) + sin(-(d*x*e + c*e)^(2/3)*b*e^(-2/3) - a))*c*e^(4/3)/b^2)/d