Integrand size = 27, antiderivative size = 134 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=-\frac {2^{\frac {1}{2}+m} a (c+d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2}-m,-\frac {3}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m} \sqrt {c+d \sin (e+f x)}}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \] Output:
-2^(1/2+m)*a*(c+d)*AppellF1(1/2,-3/2,1/2-m,3/2,d*(1-sin(f*x+e))/(c+d),1/2- 1/2*sin(f*x+e))*cos(f*x+e)*(1+sin(f*x+e))^(1/2-m)*(a+a*sin(f*x+e))^(-1+m)* (c+d*sin(f*x+e))^(1/2)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx \] Input:
Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(3/2),x]
Output:
Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(3/2), x]
Time = 0.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 3267, 157, 27, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 3267 |
\(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {a^2 (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) \sqrt {c+d \sin (e+f x)} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\sqrt {2} a (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},-\frac {3}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\) |
Input:
Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(3/2),x]
Output:
(Sqrt[2]*a*(c - d)*AppellF1[1/2 + m, 1/2, -3/2, 3/2 + m, (1 + Sin[e + f*x] )/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*Sqrt[1 - Sin[e + f*x] ]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(f*(1 + 2*m)*(a - a*Sin [e + f*x])*Sqrt[(c + d*Sin[e + f*x])/(c - d)])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + 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] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d* x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m , n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[m]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
Input:
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x)
Output:
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x)
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")
Output:
integral((d*sin(f*x + e) + c)^(3/2)*(a*sin(f*x + e) + a)^m, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate((d*sin(f*x + e) + c)^(3/2)*(a*sin(f*x + e) + a)^m, x)
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")
Output:
integrate((d*sin(f*x + e) + c)^(3/2)*(a*sin(f*x + e) + a)^m, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:
int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(3/2),x)
Output:
int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(3/2), x)
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx=\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right ) d +\left (\int \sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m}d x \right ) c \] Input:
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2),x)
Output:
int(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m*sin(e + f*x),x)*d + i nt(sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m,x)*c