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

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

[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))/(c+d))^
(1/3)*2^(1/2)/f/(c+d*sin(f*x+e))^(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} \[ -\frac {2 \sqrt {2} a \cos (e+f x) \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}} F_1\left (\frac {1}{2};-\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1} \sqrt [3]{c+d \sin (e+f x)}} \]

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])/(c + d))^(1/3))/(f*Sqrt[1 + Sin[e + f*x]]*(c + d*Sin[e + f*x])^(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 \frac {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 {1-x} \sqrt [3]{c+d 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]{-\frac {c+d \sin (e+f x)}{-c-d}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{\sqrt {1-x} \sqrt [3]{-\frac {c}{-c-d}-\frac {d x}{-c-d}}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)} \sqrt [3]{c+d \sin (e+f x)}}\\ &=-\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]{\frac {c+d \sin (e+f x)}{c+d}}}{f \sqrt {1+\sin (e+f x)} \sqrt [3]{c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 6.27, size = 886, normalized size = 8.28 \[ a \left (\frac {\sec (e) \left (-\frac {F_1\left (-\frac {1}{3};-\frac {1}{2},-\frac {1}{2};\frac {2}{3};-\frac {\csc (e) \left (c+d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} \sin (e)\right )}{d \sqrt {\cot ^2(e)+1} \left (1-\frac {c \csc (e)}{d \sqrt {\cot ^2(e)+1}}\right )},-\frac {\csc (e) \left (c+d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} \sin (e)\right )}{d \sqrt {\cot ^2(e)+1} \left (-\frac {c \csc (e)}{d \sqrt {\cot ^2(e)+1}}-1\right )}\right ) \cot (e) \sin \left (f x-\tan ^{-1}(\cot (e))\right )}{\sqrt {\cot ^2(e)+1} \sqrt {\frac {\cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} d+\sqrt {\cot ^2(e)+1} d}{d \sqrt {\cot ^2(e)+1}-c \csc (e)}} \sqrt {\frac {d \sqrt {\cot ^2(e)+1}-d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1}}{\sqrt {\cot ^2(e)+1} d+c \csc (e)}} \sqrt [3]{c+d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} \sin (e)}}-\frac {\frac {3 d \sin (e) \left (c+d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} \sin (e)\right )}{2 \left (d^2 \cos ^2(e)+d^2 \sin ^2(e)\right )}-\frac {\cot (e) \sin \left (f x-\tan ^{-1}(\cot (e))\right )}{\sqrt {\cot ^2(e)+1}}}{\sqrt [3]{c+d \cos \left (f x-\tan ^{-1}(\cot (e))\right ) \sqrt {\cot ^2(e)+1} \sin (e)}}\right ) (\sin (e+f x)+1)}{f \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}+\frac {3 (c+d \sin (e+f x))^{2/3} \tan (e) (\sin (e+f x)+1)}{2 d f \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2}+\frac {3 F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};-\frac {\sec (e) \left (c+d \cos (e) \sin \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\tan ^2(e)+1}\right )}{d \sqrt {\tan ^2(e)+1} \left (1-\frac {c \sec (e)}{d \sqrt {\tan ^2(e)+1}}\right )},-\frac {\sec (e) \left (c+d \cos (e) \sin \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\tan ^2(e)+1}\right )}{d \sqrt {\tan ^2(e)+1} \left (-\frac {c \sec (e)}{d \sqrt {\tan ^2(e)+1}}-1\right )}\right ) \sec (e) \sec \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\frac {d \sqrt {\tan ^2(e)+1}-d \sin \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\tan ^2(e)+1}}{\sqrt {\tan ^2(e)+1} d+c \sec (e)}} \sqrt {\frac {\sin \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\tan ^2(e)+1} d+\sqrt {\tan ^2(e)+1} d}{d \sqrt {\tan ^2(e)+1}-c \sec (e)}} \left (c+d \cos (e) \sin \left (f x+\tan ^{-1}(\tan (e))\right ) \sqrt {\tan ^2(e)+1}\right )^{2/3} (\sin (e+f x)+1)}{2 d f \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )^2 \sqrt {\tan ^2(e)+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*((Sec[e]*(1 + Sin[e + f*x])*(-((AppellF1[-1/3, -1/2, -1/2, 2/3, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*S
qrt[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*C
os[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 + Cot[e]^2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*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])]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + C
ot[e]^2]*Sin[e])^(1/3))) - ((3*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(2*(d^2*C
os[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])^(1/3)))/(f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (3*(1 + Sin[e + f*x]
)*(c + d*Sin[e + f*x])^(2/3)*Tan[e])/(2*d*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (3*AppellF1[2/3, 1/
2, 1/2, 5/3, -((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])^(2/3))/(2*d*f*(Cos[e/2 + (f*x)/2] +
Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}, x\right ) \]

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 \[ \int \frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}\,{d x} \]

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 = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {a +a \sin \left (f x +e \right )}{\left (c +d \sin \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx \]

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 \[ \int \frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}}\,{d x} \]

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 \[ \int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \]

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 \[ a \left (\int \frac {\sin {\left (e + f x \right )}}{\sqrt [3]{c + d \sin {\left (e + f x \right )}}}\, dx + \int \frac {1}{\sqrt [3]{c + d \sin {\left (e + f x \right )}}}\, dx\right ) \]

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

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