Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d \sqrt {a+b \sin (c+d x)}} \]
-2*e*AppellF1(-1/2,1/2-1/2*p,1/2-1/2*p,1/2,(a+b*sin(d*x+c))/(a-b),(a+b*sin (d*x+c))/(a+b))*(e*cos(d*x+c))^(-1+p)*(1+(-a-b*sin(d*x+c))/(a-b))^(1/2-1/2 *p)*(1+(-a-b*sin(d*x+c))/(a+b))^(1/2-1/2*p)/b/d/(a+b*sin(d*x+c))^(1/2)
Time = 5.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{b d \sqrt {a+b \sin (c+d x)}} \]
(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - Sqrt[b^2]), (a + b*Sin[c + d*x])/(a + Sqrt[b^2])]*(e*Cos[c + d*x])^(-1 + p )*((Sqrt[b^2] - b*Sin[c + d*x])/(a + Sqrt[b^2]))^((1 - p)/2)*((Sqrt[b^2] + b*Sin[c + d*x])/(-a + Sqrt[b^2]))^((1 - p)/2))/(b*d*Sqrt[a + b*Sin[c + d* x]])
Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3042, 3183, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3183 |
\(\displaystyle \frac {e (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \int \frac {\left (-\frac {\sin (c+d x) b}{a-b}-\frac {b}{a-b}\right )^{\frac {p-1}{2}} \left (\frac {b}{a+b}-\frac {b \sin (c+d x)}{a+b}\right )^{\frac {p-1}{2}}}{(a+b \sin (c+d x))^{3/2}}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {2 e (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {1}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d \sqrt {a+b \sin (c+d x)}}\) |
(-2*e*AppellF1[-1/2, (1 - p)/2, (1 - p)/2, 1/2, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*(e*Cos[c + d*x])^(-1 + p)*(1 - (a + b*Si n[c + d*x])/(a - b))^((1 - p)/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^((1 - p)/2))/(b*d*Sqrt[a + b*Sin[c + d*x]])
3.7.27.3.1 Defintions of rubi rules used
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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin [e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p - 1) /2))) Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*( x/(a + b)))^((p - 1)/2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && !IGtQ[m, 0]
\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral(-sqrt(b*sin(d*x + c) + a)*(e*cos(d*x + c))^p/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), x)
\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]