Integrand size = 25, antiderivative size = 121 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {6 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}} \]
6/5*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*a^2*sec(d *x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+12/5*a^2*sin(d*x+c)*sec(d* x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 a \sqrt {a (1+\cos (c+d x))} (4+3 \cos (c+d x)+3 \cos (2 (c+d x))) \sec ^{\frac {5}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{5 d} \]
(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(4 + 3*Cos[c + d*x] + 3*Cos[2*(c + d*x)])* Sec[c + d*x]^(5/2)*Tan[(c + d*x)/2])/(5*d)
Time = 0.66 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4710, 3042, 3241, 27, 3042, 3251, 3042, 3250}
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.
\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\cos (c+d x) a+a)^{3/2}}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3241 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {2}{5} a \int -\frac {9 \sqrt {\cos (c+d x) a+a}}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {9}{5} a \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {9}{5} a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
\(\Big \downarrow \) 3251 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {9}{5} a \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {9}{5} a \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )\) |
\(\Big \downarrow \) 3250 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {9}{5} a \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a^2*Sin[c + d*x])/(5*d*Cos[c + d *x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (9*a*((2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[ c + d*x]]*Sqrt[a + a*Cos[c + d*x]])))/5)
3.4.47.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 6.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {2 \cot \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sec ^{\frac {7}{2}}\left (d x +c \right )\right ) \left (6 \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right )-1\right ) a}{5 d}\) | \(63\) |
-2/5/d*cot(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)*sec(d*x+c)^(7/2)*(6*cos(d*x+c)^ 3-3*cos(d*x+c)^2-2*cos(d*x+c)-1)*a
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \, {\left (6 \, a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}} \]
2/5*(6*a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sqrt(a*cos(d*x + c) + a)*s in(d*x + c)/((d*cos(d*x + c)^3 + d*cos(d*x + c)^2)*sqrt(cos(d*x + c)))
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).
Time = 0.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.79 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{5 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} \]
4/5*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(2)*a^(3/2 )*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/( cos(d*x + c) + 1)^5 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^ 7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(2*sin(d* x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1))
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
Time = 1.70 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^{3/2} \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {4\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (c+d\,x\right )+6\,\sin \left (2\,c+2\,d\,x\right )+11\,\sin \left (3\,c+3\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+3\,\sin \left (5\,c+5\,d\,x\right )\right )}{5\,d\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]