Integrand size = 25, antiderivative size = 107 \[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=-\frac {2 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}}}{f \sqrt {1+\sin (e+f x)} \sqrt [3]{c+d \sin (e+f x)}} \] Output:
-2*2^(1/2)*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)/f/(1+sin(f*x+e))^(1/2)/( c+d*sin(f*x+e))^(1/3)
Leaf count is larger than twice the leaf count of optimal. \(886\) vs. \(2(107)=214\).
Time = 6.61 (sec) , antiderivative size = 886, normalized size of antiderivative = 8.28 \[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx =\text {Too large to display} \] Input:
Integrate[(a + a*Sin[e + f*x])/(c + d*Sin[e + f*x])^(1/3),x]
Output:
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]]]*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 + Cot[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])^(1/3))) - ((3*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(2*(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]]]*Sqr t[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*(C os[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 + Arc Tan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[f*x ...
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3234, 156, 155}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a \sin (e+f x)+a}{\sqrt [3]{c+d \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \sin (e+f x)+a}{\sqrt [3]{c+d \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3234 |
\(\displaystyle \frac {a \cos (e+f x) \int \frac {\sqrt {\sin (e+f x)+1}}{\sqrt {1-\sin (e+f x)} \sqrt [3]{c+d \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {a \cos (e+f x) \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}} \int \frac {\sqrt {\sin (e+f x)+1}}{\sqrt {1-\sin (e+f x)} \sqrt [3]{\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1} \sqrt [3]{c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {2 \sqrt {2} a \cos (e+f x) \sqrt [3]{\frac {c+d \sin (e+f x)}{c+d}} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1} \sqrt [3]{c+d \sin (e+f x)}}\) |
Input:
Int[(a + a*Sin[e + f*x])/(c + d*Sin[e + f*x])^(1/3),x]
Output:
(-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))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[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*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[c*(Cos[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sq rt[1 - Sin[e + f*x]])) Subst[Int[(a + b*x)^m*(Sqrt[1 + (d/c)*x]/Sqrt[1 - (d/c)*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && N eQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]
\[\int \frac {a +a \sin \left (f x +e \right )}{\left (c +d \sin \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Input:
int((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x)
Output:
int((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x)
\[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x, algorithm="fricas")
Output:
integral((a*sin(f*x + e) + a)/(d*sin(f*x + e) + c)^(1/3), x)
\[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=a \left (\int \frac {\sin {\left (e + f x \right )}}{\sqrt [3]{c + d \sin {\left (e + f x \right )}}}\, dx + \int \frac {1}{\sqrt [3]{c + d \sin {\left (e + f x \right )}}}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))**(1/3),x)
Output:
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))
\[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)/(d*sin(f*x + e) + c)^(1/3), x)
\[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=\int { \frac {a \sin \left (f x + e\right ) + a}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)/(d*sin(f*x + e) + c)^(1/3), x)
Timed out. \[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=\int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \] Input:
int((a + a*sin(e + f*x))/(c + d*sin(e + f*x))^(1/3),x)
Output:
int((a + a*sin(e + f*x))/(c + d*sin(e + f*x))^(1/3), x)
\[ \int \frac {a+a \sin (e+f x)}{\sqrt [3]{c+d \sin (e+f x)}} \, dx=a \left (\int \frac {\sin \left (f x +e \right )}{\left (\sin \left (f x +e \right ) d +c \right )^{\frac {1}{3}}}d x +\int \frac {1}{\left (\sin \left (f x +e \right ) d +c \right )^{\frac {1}{3}}}d x \right ) \] Input:
int((a+a*sin(f*x+e))/(c+d*sin(f*x+e))^(1/3),x)
Output:
a*(int(sin(e + f*x)/(sin(e + f*x)*d + c)**(1/3),x) + int(1/(sin(e + f*x)*d + c)**(1/3),x))