3.3.50 \(\int (c e+d e x)^{2/3} \sin (a+\frac {b}{(c+d x)^{2/3}}) \, dx\) [250]

Optimal. Leaf size=262 \[ \frac {2 b \sqrt [3]{c+d x} (e (c+d x))^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d}+\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{5 d (c+d x)^{2/3}}-\frac {4 \sqrt {2} b^{5/2} \sqrt {\pi } (e (c+d x))^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{5 d (c+d x)^{2/3}}-\frac {4 b^2 (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d \sqrt [3]{c+d x}}+\frac {3 (c+d x) (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d} \]

[Out]

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)*
FresnelC(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)

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Rubi [A]
time = 0.17, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3516, 3498, 3496, 3490, 3468, 3469, 3435, 3433, 3432} \begin {gather*} \frac {4 \sqrt {2} \sqrt {\pi } b^{5/2} \cos (a) (e (c+d x))^{2/3} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{5 d (c+d x)^{2/3}}-\frac {4 \sqrt {2} \sqrt {\pi } b^{5/2} \sin (a) (e (c+d x))^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{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}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)*Cos[a + b/(c + d*x)^(2/3)])/(5*d) + (4*Sqrt[2]*b^(5/2)*Sqrt[Pi]*(e*(c
+ d*x))^(2/3)*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/(5*d*(c + d*x)^(2/3)) - (4*Sqrt[2]*b^(5/
2)*Sqrt[Pi]*(e*(c + d*x))^(2/3)*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(5*d*(c + d*x)^(2/3)) -
(4*b^2*(e*(c + d*x))^(2/3)*Sin[a + b/(c + d*x)^(2/3)])/(5*d*(c + d*x)^(1/3)) + (3*(c + d*x)*(e*(c + d*x))^(2/
3)*Sin[a + b/(c + d*x)^(2/3)])/(5*d)

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 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 3469

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] + Dist[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 3490

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
]

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 (c e+d e x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx &=\frac {\text {Subst}\left (\int (e x)^{2/3} \sin \left (a+\frac {b}{x^{2/3}}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {(e (c+d x))^{2/3} \text {Subst}\left (\int x^{2/3} \sin \left (a+\frac {b}{x^{2/3}}\right ) \, dx,x,c+d x\right )}{d (c+d x)^{2/3}}\\ &=\frac {\left (3 (e (c+d x))^{2/3}\right ) \text {Subst}\left (\int x^4 \sin \left (a+\frac {b}{x^2}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d (c+d x)^{2/3}}\\ &=-\frac {\left (3 (e (c+d x))^{2/3}\right ) \text {Subst}\left (\int \frac {\sin \left (a+b x^2\right )}{x^6} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d (c+d x)^{2/3}}\\ &=\frac {3 (c+d x) (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d}-\frac {\left (6 b (e (c+d x))^{2/3}\right ) \text {Subst}\left (\int \frac {\cos \left (a+b x^2\right )}{x^4} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{5 d (c+d x)^{2/3}}\\ &=\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 {3 (c+d x) (e (c+d x))^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{5 d}+\frac {\left (4 b^2 (e (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 )}{5 d (c+d x)^{2/3}}\\ &=\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 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}+\frac {\left (8 b^3 (e (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 )}{5 d (c+d x)^{2/3}}\\ &=\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 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}+\frac {\left (8 b^3 (e (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 )}{5 d (c+d x)^{2/3}}-\frac {\left (8 b^3 (e (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 )}{5 d (c+d x)^{2/3}}\\ &=\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) C\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} S\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}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 228, normalized size = 0.87 \begin {gather*} \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) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-4 b^{5/2} \sqrt {2 \pi } S\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((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/Pi])/(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))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {2}{3}} \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.83, size = 823, normalized size = 3.14 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*((((-I*e^(2/3)*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*e^(2/3)*gamma(-5/2, -I*b/(d*x + c)^(2/3))
)*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (I*e^(2/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) - I*e^(2/3)
*gamma(-5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (e^(2/3)*gamma(-5/2, I*b*conjugate
((d*x + c)^(-2/3))) + e^(2/3)*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + c)) - (e^(2
/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + e^(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) - ((e^(2/3)*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + e^(2/3)*gamma(-5/2,
-I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (e^(2/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(
-2/3))) + e^(2/3)*gamma(-5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) - (-I*e^(2/3)*gamma
(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*e^(2/3)*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan
2(0, d*x + c)) - (-I*e^(2/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*e^(2/3)*gamma(-5/2, I*b/(d*x +
c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*sin(a))*d*x + (((-I*e^(2/3)*gamma(-5/2, I*b*conjugate((d*x
+ c)^(-2/3))) + I*e^(2/3)*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (I*e^(2/3
)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) - I*e^(2/3)*gamma(-5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/
3*arctan2(0, d*x + c)) + (e^(2/3)*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + e^(2/3)*gamma(-5/2, -I*b/(d*x
+ c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + c)) - (e^(2/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) +
e^(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) - ((e^(2/3)*gamma(-5
/2, I*b*conjugate((d*x + c)^(-2/3))) + e^(2/3)*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0,
d*x + c)) + (e^(2/3)*gamma(-5/2, -I*b*conjugate((d*x + c)^(-2/3))) + e^(2/3)*gamma(-5/2, I*b/(d*x + c)^(2/3)))
*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) - (-I*e^(2/3)*gamma(-5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*e^(2/3)
*gamma(-5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + c)) - (-I*e^(2/3)*gamma(-5/2, -I*b*conju
gate((d*x + c)^(-2/3))) + I*e^(2/3)*gamma(-5/2, I*b/(d*x + c)^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*
sin(a))*c)*(d*x + c)^(2/3)*(b/(d*x + c)^(2/3))^(5/2)/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((d*e*x+c*e)^(2/3)*sin(a+b/(d*x+c)^(2/3)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \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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