Integrand size = 21, antiderivative size = 186 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=-\frac {10 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{21 d}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}+\frac {10 \left (a^2+b^2\right )^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{21 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \]
-2/7*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c))^(5/2)/d-10/21 *(a^2+b^2)*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)/d +10/21*(a^2+b^2)^2*(cos(1/2*c+1/2*d*x-1/2*arctan(a,b))^2)^(1/2)/cos(1/2*c+ 1/2*d*x-1/2*arctan(a,b))*EllipticF(sin(1/2*c+1/2*d*x-1/2*arctan(a,b)),2^(1 /2))*((a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)/d/(a*cos(d*x+c)+b *sin(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.43 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {a \cos (c+d x)+b \sin (c+d x)} \left (-23 b \left (a^2+b^2\right ) \cos (c+d x)+\left (-9 a^2 b+3 b^3\right ) \cos (3 (c+d x))+2 a \left (13 a^2+7 b^2+3 \left (a^2-3 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )+\frac {20 \left (a^2+b^2\right )^2 \sqrt {\cos ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right ) \tan \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}{\sqrt {\sqrt {1+\frac {a^2}{b^2}} b \sin \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}}}{42 d} \]
(Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]*(-23*b*(a^2 + b^2)*Cos[c + d*x] + ( -9*a^2*b + 3*b^3)*Cos[3*(c + d*x)] + 2*a*(13*a^2 + 7*b^2 + 3*(a^2 - 3*b^2) *Cos[2*(c + d*x)])*Sin[c + d*x]) + (20*(a^2 + b^2)^2*Sqrt[Cos[c + d*x + Ar cTan[a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/ b]]^2]*Tan[c + d*x + ArcTan[a/b]])/Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]])/(42*d)
Time = 0.56 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3552, 3042, 3552, 3042, 3557, 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.
| \(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \cos (c+d x)+b \sin (c+d x))^{7/2}dx\) |
| \(\Big \downarrow \) 3552 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \int (a \cos (c+d x)+b \sin (c+d x))^{3/2}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \int (a \cos (c+d x)+b \sin (c+d x))^{3/2}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3552 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \left (\frac {1}{3} \left (a^2+b^2\right ) \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\right )-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \left (\frac {1}{3} \left (a^2+b^2\right ) \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}dx-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\right )-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3557 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \left (\frac {\left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}}dx}{3 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\right )-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \left (\frac {\left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\sin \left (c+d x-\tan ^{-1}(a,b)+\frac {\pi }{2}\right )}}dx}{3 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\right )-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {5}{7} \left (a^2+b^2\right ) \left (\frac {2 \left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right )}{3 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{3 d}\right )-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{5/2}}{7 d}\) |
(-2*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(5 /2))/(7*d) + (5*(a^2 + b^2)*((-2*(b*Cos[c + d*x] - a*Sin[c + d*x])*Sqrt[a* Cos[c + d*x] + b*Sin[c + d*x]])/(3*d) + (2*(a^2 + b^2)*EllipticF[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[(a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^ 2]])/(3*d*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])))/7
3.3.32.3.1 Defintions of rubi rules used
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[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(-(b*Cos[c + d*x] - a*Sin[c + d*x]))*((a*Cos[c + d*x] + b* Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[(n - 1)*((a^2 + b^2)/n) Int[(a*Co s[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && N eQ[a^2 + b^2, 0] && !IntegerQ[(n - 1)/2] && GtQ[n, 1]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S in[c + d*x])/Sqrt[a^2 + b^2])^n Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])
Time = 1.36 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.99
| method | result | size |
| default | \(-\frac {\left (a^{2}+b^{2}\right )^{2} \left (-6 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{5}+5 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-4 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{3}+10 \sin \left (d x +c -\arctan \left (-a , b\right )\right )\right )}{21 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\) | \(185\) |
-1/21*(a^2+b^2)^2*(-6*sin(d*x+c-arctan(-a,b))^5+5*(-sin(d*x+c-arctan(-a,b) )+1)^(1/2)*(2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*sin(d*x+c-arctan(-a,b))^(1/ 2)*EllipticF((-sin(d*x+c-arctan(-a,b))+1)^(1/2),1/2*2^(1/2))-4*sin(d*x+c-a rctan(-a,b))^3+10*sin(d*x+c-arctan(-a,b)))/cos(d*x+c-arctan(-a,b))/(sin(d* x+c-arctan(-a,b))*(a^2+b^2)^(1/2))^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.40 \[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} - a^{2} b + i \, a b^{2} - b^{3}\right )} \sqrt {a - i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} - a^{2} b - i \, a b^{2} - b^{3}\right )} \sqrt {a + i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} b - 8 \, b^{3}\right )} \cos \left (d x + c\right ) - {\left (5 \, a^{3} + 8 \, a b^{2} + 3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}{21 \, d} \]
-1/21*(5*sqrt(2)*(I*a^3 - a^2*b + I*a*b^2 - b^3)*sqrt(a - I*b)*weierstrass PInverse(-4*(a^2 + 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*a^3 - a^2*b - I*a*b^2 - b^3)*sqrt(a + I*b)*weierstr assPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) - I*sin( d*x + c)) + 2*(3*(3*a^2*b - b^3)*cos(d*x + c)^3 - (a^2*b - 8*b^3)*cos(d*x + c) - (5*a^3 + 8*a*b^2 + 3*(a^3 - 3*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))* sqrt(a*cos(d*x + c) + b*sin(d*x + c)))/d
Timed out. \[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}} \,d x } \]
\[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=\int { {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (a \cos (c+d x)+b \sin (c+d x))^{7/2} \, dx=\int {\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{7/2} \,d x \]