Integrand size = 25, antiderivative size = 165 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a b}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}} \]
-2/5*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))/d/e/(e*sin(d*x+c))^(5/2)-2/5*a*b/d/ e^3/(e*sin(d*x+c))^(1/2)-2/5*(3*a^2-2*b^2)*cos(d*x+c)/d/e^3/(e*sin(d*x+c)) ^(1/2)+2/5*(3*a^2-2*b^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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^( 1/2)/d/e^4/sin(d*x+c)^(1/2)
Time = 1.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {8 a b+\left (7 a^2+2 b^2\right ) \cos (c+d x)-3 a^2 \cos (3 (c+d x))+2 b^2 \cos (3 (c+d x))-4 \left (3 a^2-2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {5}{2}}(c+d x)}{10 d e (e \sin (c+d x))^{5/2}} \]
-1/10*(8*a*b + (7*a^2 + 2*b^2)*Cos[c + d*x] - 3*a^2*Cos[3*(c + d*x)] + 2*b ^2*Cos[3*(c + d*x)] - 4*(3*a^2 - 2*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2 ]*Sin[c + d*x]^(5/2))/(d*e*(e*Sin[c + d*x])^(5/2))
Time = 0.71 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3170, 27, 3042, 3148, 3042, 3116, 3042, 3121, 3042, 3119}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle -\frac {2 \int -\frac {3 a^2+b \cos (c+d x) a-2 b^2}{2 (e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 a^2+b \cos (c+d x) a-2 b^2}{(e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a^2-b \sin \left (c+d x-\frac {\pi }{2}\right ) a-2 b^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(e \sin (c+d x))^{3/2}}dx-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \int \frac {1}{(e \sin (c+d x))^{3/2}}dx-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\int \sqrt {e \sin (c+d x)}dx}{e^2}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\left (3 a^2-2 b^2\right ) \left (-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )-\frac {2 a b}{d e \sqrt {e \sin (c+d x)}}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}\) |
(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x]))/(5*d*e*(e*Sin[c + d*x])^(5/ 2)) + ((-2*a*b)/(d*e*Sqrt[e*Sin[c + d*x]]) + (3*a^2 - 2*b^2)*((-2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqr t[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]])))/(5*e^2)
3.1.48.3.1 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[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x ])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g }, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* p] || IntegerQ[m])
Time = 3.50 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.13
| method | result | size |
| default | \(\frac {-\frac {4 a b}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-4 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 a^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(351\) |
| parts | \(\frac {a^{2} \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{5}\left (d x +c \right )\right )-4 \left (\sin ^{3}\left (d x +c \right )\right )-2 \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{2} \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 \left (\sin ^{5}\left (d x +c \right )\right )-3 \left (\sin ^{3}\left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {4 a b}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}} e d}\) | \(360\) |
(-4/5*a*b/e/(e*sin(d*x+c))^(5/2)+1/5/e^3*(6*(1-sin(d*x+c))^(1/2)*(2*sin(d* x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2)) *a^2-4*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*Ellipt icE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-3*(1-sin(d*x+c))^(1/2)*(2*sin(d* x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2)) *a^2+2*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*Ellipt icF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+6*a^2*cos(d*x+c)^4*sin(d*x+c)-4* b^2*cos(d*x+c)^4*sin(d*x+c)-8*a^2*cos(d*x+c)^2*sin(d*x+c)+2*b^2*cos(d*x+c) ^2*sin(d*x+c))/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\frac {{\left (\sqrt {2} {\left (-3 i \, a^{2} + 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (3 i \, a^{2} - 2 i \, b^{2}\right )}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (3 i \, a^{2} - 2 i \, b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-3 i \, a^{2} + 2 i \, b^{2}\right )}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left ({\left (3 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, a b - {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4}\right )} \sin \left (d x + c\right )} \]
1/5*((sqrt(2)*(-3*I*a^2 + 2*I*b^2)*cos(d*x + c)^2 + sqrt(2)*(3*I*a^2 - 2*I *b^2))*sqrt(-I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4 , 0, cos(d*x + c) + I*sin(d*x + c))) + (sqrt(2)*(3*I*a^2 - 2*I*b^2)*cos(d* x + c)^2 + sqrt(2)*(-3*I*a^2 + 2*I*b^2))*sqrt(I*e)*sin(d*x + c)*weierstras sZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2* ((3*a^2 - 2*b^2)*cos(d*x + c)^3 - 2*a*b - (4*a^2 - b^2)*cos(d*x + c))*sqrt (e*sin(d*x + c)))/((d*e^4*cos(d*x + c)^2 - d*e^4)*sin(d*x + c))
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]