Integrand size = 25, antiderivative size = 85 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=-\frac {2^{\frac {9}{4}+m} (e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {5}{4}-m} (a+a \sin (c+d x))^m}{5 d e} \] Output:
-1/5*2^(9/4+m)*(e*cos(d*x+c))^(5/2)*hypergeom([5/4, -1/4-m],[9/4],1/2-1/2* sin(d*x+c))*(1+sin(d*x+c))^(-5/4-m)*(a+a*sin(d*x+c))^m/d/e
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=-\frac {2^{\frac {9}{4}+m} (e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-m,\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {5}{4}-m} (a (1+\sin (c+d x)))^m}{5 d e} \] Input:
Integrate[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^m,x]
Output:
-1/5*(2^(9/4 + m)*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, -1/4 - m, 9/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(-5/4 - m)*(a*(1 + Sin[c + d *x]))^m)/(d*e)
Time = 0.30 (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.
\(\displaystyle \int (e \cos (c+d x))^{3/2} (a \sin (c+d x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (e \cos (c+d x))^{3/2} (a \sin (c+d x)+a)^mdx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 (e \cos (c+d x))^{5/2} \int \sqrt [4]{a-a \sin (c+d x)} (\sin (c+d x) a+a)^{m+\frac {1}{4}}d\sin (c+d x)}{d e (a-a \sin (c+d x))^{5/4} (a \sin (c+d x)+a)^{5/4}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{m+\frac {1}{4}} (e \cos (c+d x))^{5/2} (\sin (c+d x)+1)^{-m-\frac {1}{4}} (a \sin (c+d x)+a)^{m-1} \int \left (\frac {1}{2} \sin (c+d x)+\frac {1}{2}\right )^{m+\frac {1}{4}} \sqrt [4]{a-a \sin (c+d x)}d\sin (c+d x)}{d e (a-a \sin (c+d x))^{5/4}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {a 2^{m+\frac {9}{4}} (e \cos (c+d x))^{5/2} (\sin (c+d x)+1)^{-m-\frac {1}{4}} (a \sin (c+d x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-m-\frac {1}{4},\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e}\) |
Input:
Int[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^m,x]
Output:
-1/5*(2^(9/4 + m)*a*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, -1/4 - m , 9/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(-1/4 - m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e)
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]
\[\int \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
Input:
int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x)
Output:
int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x)
\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
Output:
integral(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^m*e*cos(d*x + c), x)
Timed out. \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\text {Timed out} \] Input:
integrate((e*cos(d*x+c))**(3/2)*(a+a*sin(d*x+c))**m,x)
Output:
Timed out
\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
Output:
integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^m, x)
\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
Output:
integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^m, x)
Timed out. \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m \,d x \] Input:
int((e*cos(c + d*x))^(3/2)*(a + a*sin(c + d*x))^m,x)
Output:
int((e*cos(c + d*x))^(3/2)*(a + a*sin(c + d*x))^m, x)
\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^m \, dx=\sqrt {e}\, \left (\int \left (\sin \left (d x +c \right ) a +a \right )^{m} \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) e \] Input:
int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^m,x)
Output:
sqrt(e)*int((sin(c + d*x)*a + a)**m*sqrt(cos(c + d*x))*cos(c + d*x),x)*e