Integrand size = 25, antiderivative size = 154 \[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {2 \left (7 a^2+2 b^2\right ) e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {2 \left (7 a^2+2 b^2\right ) e \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}+\frac {18 a b (e \sin (c+d x))^{5/2}}{35 d e}+\frac {2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{7 d e} \]
18/35*a*b*(e*sin(d*x+c))^(5/2)/d/e+2/7*b*(a+b*cos(d*x+c))*(e*sin(d*x+c))^( 5/2)/d/e-2/21*(7*a^2+2*b^2)*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/ 2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c )^(1/2)/d/(e*sin(d*x+c))^(1/2)-2/21*(7*a^2+2*b^2)*e*cos(d*x+c)*(e*sin(d*x+ c))^(1/2)/d
Time = 1.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {\left (-\frac {1}{2} \left (5 \left (28 a^2+5 b^2\right ) \cos (c+d x)+3 b (-28 a+28 a \cos (2 (c+d x))+5 b \cos (3 (c+d x)))\right ) \csc (c+d x)-\frac {10 \left (7 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )}{\sin ^{\frac {3}{2}}(c+d x)}\right ) (e \sin (c+d x))^{3/2}}{105 d} \]
((-1/2*((5*(28*a^2 + 5*b^2)*Cos[c + d*x] + 3*b*(-28*a + 28*a*Cos[2*(c + d* x)] + 5*b*Cos[3*(c + d*x)]))*Csc[c + d*x]) - (10*(7*a^2 + 2*b^2)*EllipticF [(-2*c + Pi - 2*d*x)/4, 2])/Sin[c + d*x]^(3/2))*(e*Sin[c + d*x])^(3/2))/(1 05*d)
Time = 0.66 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3171, 27, 3042, 3148, 3042, 3115, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int (e \sin (c+d x))^{3/2} (a+b \cos (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 3171 |
| \(\displaystyle \frac {2}{7} \int \frac {1}{2} \left (7 a^2+9 b \cos (c+d x) a+2 b^2\right ) (e \sin (c+d x))^{3/2}dx+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \left (7 a^2+9 b \cos (c+d x) a+2 b^2\right ) (e \sin (c+d x))^{3/2}dx+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2} \left (7 a^2-9 b \sin \left (c+d x-\frac {\pi }{2}\right ) a+2 b^2\right )dx+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \int (e \sin (c+d x))^{3/2}dx+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \int (e \sin (c+d x))^{3/2}dx+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{7} \left (\left (7 a^2+2 b^2\right ) \left (\frac {2 e^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {18 a b (e \sin (c+d x))^{5/2}}{5 d e}\right )+\frac {2 b (e \sin (c+d x))^{5/2} (a+b \cos (c+d x))}{7 d e}\) |
(2*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2))/(7*d*e) + ((18*a*b*(e*Si n[c + d*x])^(5/2))/(5*d*e) + (7*a^2 + 2*b^2)*((2*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d)))/7
3.1.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[1/(m + p) Int[(g*Cos[e + f*x])^p* (a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1) *Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m])
Time = 3.14 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.49
| method | result | size |
| default | \(-\frac {e^{2} \left (30 b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+35 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+10 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}+84 a b \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+70 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-84 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{105 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(229\) |
| parts | \(-\frac {a^{2} e^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{2} e^{2} \left (3 \left (\sin ^{5}\left (d x +c \right )\right )+\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-5 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{21 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {4 a b \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}\) | \(230\) |
-1/105/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^2*(30*b^2*cos(d*x+c)^4*sin(d*x+c) +35*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF ((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+10*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+ c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b ^2+84*a*b*cos(d*x+c)^3*sin(d*x+c)+70*a^2*cos(d*x+c)^2*sin(d*x+c)-10*b^2*co s(d*x+c)^2*sin(d*x+c)-84*a*b*cos(d*x+c)*sin(d*x+c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.98 \[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {5 \, \sqrt {2} {\left (7 \, a^{2} + 2 \, b^{2}\right )} \sqrt {-i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (7 \, a^{2} + 2 \, b^{2}\right )} \sqrt {i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (15 \, b^{2} e \cos \left (d x + c\right )^{3} + 42 \, a b e \cos \left (d x + c\right )^{2} - 42 \, a b e + 5 \, {\left (7 \, a^{2} - b^{2}\right )} e \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}}{105 \, d} \]
1/105*(5*sqrt(2)*(7*a^2 + 2*b^2)*sqrt(-I*e)*e*weierstrassPInverse(4, 0, co s(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(7*a^2 + 2*b^2)*sqrt(I*e)*e*weier strassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(15*b^2*e*cos(d*x + c)^3 + 42*a*b*e*cos(d*x + c)^2 - 42*a*b*e + 5*(7*a^2 - b^2)*e*cos(d*x + c))*sqrt(e*sin(d*x + c)))/d
\[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int \left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (a + b \cos {\left (c + d x \right )}\right )^{2}\, dx \]
\[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
\[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]