Integrand size = 25, antiderivative size = 85 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {9}{4}+m} (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a+a \sin (e+f x))^m}{5 f g} \] Output:
-1/5*2^(9/4+m)*(g*cos(f*x+e))^(5/2)*hypergeom([5/4, -1/4-m],[9/4],1/2-1/2* sin(f*x+e))*(1+sin(f*x+e))^(-5/4-m)*(a+a*sin(f*x+e))^m/f/g
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {9}{4}+m} (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a (1+\sin (e+f x)))^m}{5 f g} \] Input:
Integrate[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m,x]
Output:
-1/5*(2^(9/4 + m)*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[5/4, -1/4 - m, 9/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-5/4 - m)*(a*(1 + Sin[e + f *x]))^m)/(f*g)
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3168, 80, 79}
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.
\(\displaystyle \int (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 (g \cos (e+f x))^{5/2} \int \sqrt [4]{a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{4}}d\sin (e+f x)}{f g (a-a \sin (e+f x))^{5/4} (a \sin (e+f x)+a)^{5/4}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{m+\frac {1}{4}} (g \cos (e+f x))^{5/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-1} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{4}} \sqrt [4]{a-a \sin (e+f x)}d\sin (e+f x)}{f g (a-a \sin (e+f x))^{5/4}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {a 2^{m+\frac {9}{4}} (g \cos (e+f x))^{5/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-m-\frac {1}{4},\frac {9}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f g}\) |
Input:
Int[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m,x]
Output:
-1/5*(2^(9/4 + m)*a*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[5/4, -1/4 - m , 9/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(-1 + m))/(f*g)
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((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, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin [e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))) Subst[Int[(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
Input:
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x)
Output:
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x)
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")
Output:
integral(sqrt(g*cos(f*x + e))*(a*sin(f*x + e) + a)^m*g*cos(f*x + e), x)
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\text {Timed out} \] Input:
integrate((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**m,x)
Output:
Timed out
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")
Output:
integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m, x)
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x, algorithm="giac")
Output:
integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m, x)
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \] Input:
int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^m,x)
Output:
int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^m, x)
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m \, dx=\sqrt {g}\, \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right )d x \right ) g \] Input:
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m,x)
Output:
sqrt(g)*int((sin(e + f*x)*a + a)**m*sqrt(cos(e + f*x))*cos(e + f*x),x)*g