Integrand size = 25, antiderivative size = 156 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {7}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} (a+b \sin (c+d x))^{5/2} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{5 b d} \]
2/5*e*AppellF1(5/2,1/2-1/2*p,1/2-1/2*p,7/2,(a+b*sin(d*x+c))/(a-b),(a+b*sin (d*x+c))/(a+b))*(e*cos(d*x+c))^(-1+p)*(a+b*sin(d*x+c))^(5/2)*(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
Time = 3.23 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.20 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {7}{2},\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} (a+b \sin (c+d x))^{5/2} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{5 b d} \]
(2*e*AppellF1[5/2, (1 - p)/2, (1 - p)/2, 7/2, (a + b*Sin[c + d*x])/(a - Sq rt[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)*(a + b*Sin[c + d*x])^(5/2)*((Sqrt[b^2] + b*Sin[c + d*x])/(-a + Sqrt[b^2]))^((1 - p)/2))/( 5*b*d)
Time = 0.32 (sec) , antiderivative size = 156, 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.
\(\displaystyle \int (a+b \sin (c+d x))^{3/2} (e \cos (c+d x))^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (c+d x))^{3/2} (e \cos (c+d x))^pdx\) |
\(\Big \downarrow \) 3183 |
\(\displaystyle \frac {e (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \int (a+b \sin (c+d x))^{3/2} \left (-\frac {\sin (c+d x) b}{a-b}-\frac {b}{a-b}\right )^{\frac {p-1}{2}} \left (\frac {b}{a+b}-\frac {b \sin (c+d x)}{a+b}\right )^{\frac {p-1}{2}}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {2 e (a+b \sin (c+d x))^{5/2} (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {7}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{5 b d}\) |
(2*e*AppellF1[5/2, (1 - p)/2, (1 - p)/2, 7/2, (a + b*Sin[c + d*x])/(a - b) , (a + b*Sin[c + d*x])/(a + b)]*(e*Cos[c + d*x])^(-1 + p)*(a + b*Sin[c + d *x])^(5/2)*(1 - (a + b*Sin[c + d*x])/(a - b))^((1 - p)/2)*(1 - (a + b*Sin[ c + d*x])/(a + b))^((1 - p)/2))/(5*b*d)
3.7.24.3.1 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]
\[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}d x\]
\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{p} \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]
\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
Timed out. \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]