Integrand size = 21, antiderivative size = 352 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{1+n}}{b^2 f (2+n) (3+n)}-\frac {\cos (e+f x) \sin (e+f x) (a+b \sin (e+f x))^{1+n}}{b f (3+n)}-\frac {\sqrt {2} \left (2 a^2+b^2 (2+n)^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^{1+n} \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-1-n}}{b^3 f (2+n) (3+n) \sqrt {1+\sin (e+f x)}}+\frac {\sqrt {2} a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n}}{b^3 f (2+n) (3+n) \sqrt {1+\sin (e+f x)}} \] Output:
2*a*cos(f*x+e)*(a+b*sin(f*x+e))^(1+n)/b^2/f/(2+n)/(3+n)-cos(f*x+e)*sin(f*x +e)*(a+b*sin(f*x+e))^(1+n)/b/f/(3+n)-2^(1/2)*(2*a^2+b^2*(2+n)^2)*AppellF1( 1/2,-1-n,1/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+ b*sin(f*x+e))^(1+n)*((a+b*sin(f*x+e))/(a+b))^(-1-n)/b^3/f/(2+n)/(3+n)/(1+s in(f*x+e))^(1/2)+2^(1/2)*a*(2*a^2+b^2*(n^2+5*n+4))*AppellF1(1/2,-n,1/2,3/2 ,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^n/ b^3/f/(2+n)/(3+n)/(1+sin(f*x+e))^(1/2)/(((a+b*sin(f*x+e))/(a+b))^n)
\[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx \] Input:
Integrate[Sin[e + f*x]^3*(a + b*Sin[e + f*x])^n,x]
Output:
Integrate[Sin[e + f*x]^3*(a + b*Sin[e + f*x])^n, x]
Time = 0.95 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3272, 3042, 3502, 25, 3042, 3235, 3042, 3144, 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.
\(\displaystyle \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^3 (a+b \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3272 |
\(\displaystyle \frac {\int (a+b \sin (e+f x))^n \left (-2 a \sin ^2(e+f x)+b (n+2) \sin (e+f x)+a\right )dx}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (a+b \sin (e+f x))^n \left (-2 a \sin (e+f x)^2+b (n+2) \sin (e+f x)+a\right )dx}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {\int -(a+b \sin (e+f x))^n \left (a b n-\left (2 a^2+b^2 (n+2)^2\right ) \sin (e+f x)\right )dx}{b (n+2)}+\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\int (a+b \sin (e+f x))^n \left (a b n-\left (2 a^2+b^2 (n+2)^2\right ) \sin (e+f x)\right )dx}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\int (a+b \sin (e+f x))^n \left (a b n-\left (2 a^2+b^2 (n+2)^2\right ) \sin (e+f x)\right )dx}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\frac {a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \int (a+b \sin (e+f x))^ndx}{b}-\frac {\left (2 a^2+b^2 (n+2)^2\right ) \int (a+b \sin (e+f x))^{n+1}dx}{b}}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\frac {a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \int (a+b \sin (e+f x))^ndx}{b}-\frac {\left (2 a^2+b^2 (n+2)^2\right ) \int (a+b \sin (e+f x))^{n+1}dx}{b}}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\frac {a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (e+f x) \int \frac {(a+b \sin (e+f x))^n}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {\left (2 a^2+b^2 (n+2)^2\right ) \cos (e+f x) \int \frac {(a+b \sin (e+f x))^{n+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\frac {a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^n}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {(a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^{n+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\frac {2 a \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+2)}-\frac {\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^n \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}}{b (n+2)}}{b (n+3)}-\frac {\sin (e+f x) \cos (e+f x) (a+b \sin (e+f x))^{n+1}}{b f (n+3)}\) |
Input:
Int[Sin[e + f*x]^3*(a + b*Sin[e + f*x])^n,x]
Output:
-((Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x])^(1 + n))/(b*f*(3 + n))) + ((2*a*Cos[e + f*x]*(a + b*Sin[e + f*x])^(1 + n))/(b*f*(2 + n)) - ((Sqrt[ 2]*(a + b)*(2*a^2 + b^2*(2 + n)^2)*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Si n[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^n)/(b*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^n) - (Sqrt[2]*a*(2*a^2 + b^2*(4 + 5*n + n^2))*AppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^n)/(b*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^n)) /(b*(2 + n)))/(b*(3 + n))
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]]) Subst[Int[(a + b*x )^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b Int[(a + b*Sin[e + f*x])^m, x], x] + Simp[d/b Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
\[\int \sin \left (f x +e \right )^{3} \left (a +b \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x)
Output:
int(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x)
\[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{3} \,d x } \] Input:
integrate(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x, algorithm="fricas")
Output:
integral(-(cos(f*x + e)^2 - 1)*(b*sin(f*x + e) + a)^n*sin(f*x + e), x)
Timed out. \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\text {Timed out} \] Input:
integrate(sin(f*x+e)**3*(a+b*sin(f*x+e))**n,x)
Output:
Timed out
\[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{3} \,d x } \] Input:
integrate(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x, algorithm="maxima")
Output:
integrate((b*sin(f*x + e) + a)^n*sin(f*x + e)^3, x)
\[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{n} \sin \left (f x + e\right )^{3} \,d x } \] Input:
integrate(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x, algorithm="giac")
Output:
integrate((b*sin(f*x + e) + a)^n*sin(f*x + e)^3, x)
Timed out. \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int(sin(e + f*x)^3*(a + b*sin(e + f*x))^n,x)
Output:
int(sin(e + f*x)^3*(a + b*sin(e + f*x))^n, x)
\[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^n \, dx=\int \left (\sin \left (f x +e \right ) b +a \right )^{n} \sin \left (f x +e \right )^{3}d x \] Input:
int(sin(f*x+e)^3*(a+b*sin(f*x+e))^n,x)
Output:
int((sin(e + f*x)*b + a)**n*sin(e + f*x)**3,x)