Integrand size = 25, antiderivative size = 100 \[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2^{1+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (-\frac {p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-2-p)} \sqrt {a+a \sin (c+d x)}}{d e (1+p)} \] Output:
-2^(1+1/2*p)*(e*cos(d*x+c))^(p+1)*hypergeom([-1/2*p, 1/2*p+1/2],[3/2+1/2*p ],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(-1-1/2*p)*(a+a*sin(d*x+c))^(1/2)/d/e /(p+1)
Time = 0.40 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98 \[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=-\frac {2^{1+\frac {p}{2}} a \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (-\frac {p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{d (1+p) \sqrt {a (1+\sin (c+d x))}} \] Input:
Integrate[(e*Cos[c + d*x])^p*Sqrt[a + a*Sin[c + d*x]],x]
Output:
-((2^(1 + p/2)*a*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[-1/2*p, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d*x])/2])/(d*(1 + p)*(1 + Sin[c + d*x] )^(p/2)*Sqrt[a*(1 + Sin[c + d*x])]))
Time = 0.32 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.49, 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.
\(\displaystyle \int \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^pdx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \int (a-a \sin (c+d x))^{\frac {p-1}{2}} (\sin (c+d x) a+a)^{p/2}d\sin (c+d x)}{d e}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{\frac {p-1}{2}} (1-\sin (c+d x))^{\frac {1-p}{2}} (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)+\frac {p-1}{2}} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \int \left (\frac {1}{2}-\frac {1}{2} \sin (c+d x)\right )^{\frac {p-1}{2}} (\sin (c+d x) a+a)^{p/2}d\sin (c+d x)}{d e}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {a 2^{\frac {p-1}{2}+1} (1-\sin (c+d x))^{\frac {1-p}{2}} (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)+\frac {p-1}{2}} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)+\frac {p+2}{2}} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1-p}{2},\frac {p+2}{2},\frac {p+4}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{d e (p+2)}\) |
Input:
Int[(e*Cos[c + d*x])^p*Sqrt[a + a*Sin[c + d*x]],x]
Output:
(2^(1 + (-1 + p)/2)*a*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(1 - p)/2 , (2 + p)/2, (4 + p)/2, (1 + Sin[c + d*x])/2]*(1 - Sin[c + d*x])^((1 - p)/ 2)*(a - a*Sin[c + d*x])^((-1 - p)/2 + (-1 + p)/2)*(a + a*Sin[c + d*x])^((- 1 - p)/2 + (2 + p)/2))/(d*e*(2 + p))
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]
\[\int \left (e \cos \left (d x +c \right )\right )^{p} \sqrt {a +a \sin \left (d x +c \right )}d x\]
Input:
int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x)
Output:
int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x)
\[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:
integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)
\[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \] Input:
integrate((e*cos(d*x+c))**p*(a+a*sin(d*x+c))**(1/2),x)
Output:
Integral(sqrt(a*(sin(c + d*x) + 1))*(e*cos(c + d*x))**p, x)
\[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:
integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)
\[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:
integrate((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(a*sin(d*x + c) + a)*(e*cos(d*x + c))^p, x)
Timed out. \[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,\sqrt {a+a\,\sin \left (c+d\,x\right )} \,d x \] Input:
int((e*cos(c + d*x))^p*(a + a*sin(c + d*x))^(1/2),x)
Output:
int((e*cos(c + d*x))^p*(a + a*sin(c + d*x))^(1/2), x)
\[ \int (e \cos (c+d x))^p \sqrt {a+a \sin (c+d x)} \, dx=e^{p} \sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{p}d x \right ) \] Input:
int((e*cos(d*x+c))^p*(a+a*sin(d*x+c))^(1/2),x)
Output:
e**p*sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**p,x)