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3.668 \(\int (a+a \sin (e+f x)) \sqrt [3]{c+d \sin (e+f x)} \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 \sqrt {2} a \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac {1}{2};-\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}} \]

[Out]

-2*a*AppellF1(1/2,-1/3,-1/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^(1/3)*2
^(1/2)/f/((c+d*sin(f*x+e))/(c+d))^(1/3)/(1+sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2755, 139, 138} \[ -\frac {2 \sqrt {2} a \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)} F_1\left (\frac {1}{2};-\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^(1/3),x]

[Out]

(-2*Sqrt[2]*a*AppellF1[1/2, -1/2, -1/3, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x
]*(c + d*Sin[e + f*x])^(1/3))/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^(1/3))

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2755

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*C
os[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]]), Subst[Int[((a + b*x)^m*Sqrt[1 + (d*x)/c])/Sqrt
[1 - (d*x)/c], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b
^2, 0] &&  !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x)) \sqrt [3]{c+d \sin (e+f x)} \, dx &=\frac {(a \cos (e+f x)) \operatorname {Subst}\left (\int \frac {\sqrt {1+x} \sqrt [3]{c+d x}}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=\frac {\left (a \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+x} \sqrt [3]{-\frac {c}{-c-d}-\frac {d x}{-c-d}}}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)} \sqrt [3]{-\frac {c+d \sin (e+f x)}{-c-d}}}\\ &=-\frac {2 \sqrt {2} a F_1\left (\frac {1}{2};-\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{c+d \sin (e+f x)}}{f \sqrt {1+\sin (e+f x)} \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}\\ \end {align*}

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Mathematica [B]  time = 6.43, size = 1736, normalized size = 16.22 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^(1/3),x]

[Out]

a*((c*Sec[e]*(1 + Sin[e + f*x])*(-((AppellF1[-2/3, -1/2, -1/2, 1/3, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]
*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d
*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[
e]^2]))))]*Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcT
an[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - Ar
cTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 +
 Cot[e]^2]*Sin[e])^(2/3))) - ((3*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Co
s[e]^2 + d^2*Sin[e]^2) - (Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/(c + d*Cos[f*x - ArcTan[Cot[e]
]]*Sqrt[1 + Cot[e]^2]*Sin[e])^(2/3)))/(4*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (d*Sec[e]*(1 + Sin[e
 + f*x])*(-((AppellF1[-2/3, -1/2, -1/2, 1/3, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin
[e]))/(d*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e
]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2]))))]*Cot[e]*Sin[f
*x - ArcTan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Co
t[e]^2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 +
Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e])^(2/3
))) - ((3*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Cos[e]^2 + d^2*Sin[e]^2)
- (Cot[e]*Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*S
in[e])^(2/3)))/(f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + ((1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^(1/3
)*((-3*Cos[e]*Cos[f*x])/(4*f) + (3*Sin[e]*Sin[f*x])/(4*f) + (3*(c + 4*d)*Tan[e])/(4*d*f)))/(Cos[e/2 + (f*x)/2]
 + Sin[e/2 + (f*x)/2])^2 + (3*AppellF1[1/3, 1/2, 1/2, 4/3, -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*S
qrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*S
in[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))
]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]
*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Tan[e]^2])]*Sqrt[(d*Sqrt[1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]
]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1 + Tan[e]^2])]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 +
Tan[e]^2])^(1/3))/(4*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (3*c*AppellF1[1/3, 1/
2, 1/2, 4/3, -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 -
 (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/
(d*Sqrt[1 + Tan[e]^2]*(-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e
+ f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Ta
n[e]^2])]*Sqrt[(d*Sqrt[1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1
 + Tan[e]^2])]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])^(1/3))/(d*f*(Cos[e/2 + (f*x)/2] + S
in[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(1/3),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(1/3), x)

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{\frac {1}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(1/3),x)

[Out]

int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(1/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^(1/3),x)

[Out]

int((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^(1/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sqrt [3]{c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int \sqrt [3]{c + d \sin {\left (e + f x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))**(1/3),x)

[Out]

a*(Integral((c + d*sin(e + f*x))**(1/3)*sin(e + f*x), x) + Integral((c + d*sin(e + f*x))**(1/3), x))

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