Integrand size = 25, antiderivative size = 81 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {10 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:
2/3*a^2*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+10/3*a^2*sin( d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a \sqrt {a (1+\cos (c+d x))} (1+5 \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[(a + a*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(5/2),x]
Output:
(2*a*Sqrt[a*(1 + Cos[c + d*x])]*(1 + 5*Cos[c + d*x])*Tan[(c + d*x)/2])/(3* d*Cos[c + d*x]^(3/2))
Time = 0.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3241, 27, 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.
\(\displaystyle \int \frac {(a \cos (c+d x)+a)^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3241 |
\(\displaystyle \frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {2}{3} a \int -\frac {5 \sqrt {\cos (c+d x) a+a}}{2 \cos ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{3} a \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{3} a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 3250 |
\(\displaystyle \frac {2 a^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {10 a^2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\) |
Input:
Int[(a + a*Cos[c + d*x])^(3/2)/Cos[c + d*x]^(5/2),x]
Output:
(2*a^2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + ( 10*a^2*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])
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]
Time = 5.76 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {\sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (5 \sin \left (2 d x +2 c \right )+2 \sin \left (d x +c \right )\right ) a}{3 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )+1\right )}\) | \(57\) |
Input:
int((a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/d*(a*(cos(d*x+c)+1))^(1/2)*(5*sin(2*d*x+2*c)+2*sin(d*x+c))/cos(d*x+c)^ (3/2)/(cos(d*x+c)+1)*a
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \, {\left (5 \, a \cos \left (d x + c\right ) + a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")
Output:
2/3*(5*a*cos(d*x + c) + a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin (d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
\[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+a*cos(d*x+c))**(3/2)/cos(d*x+c)**(5/2),x)
Output:
Integral((a*(cos(c + d*x) + 1))**(3/2)/cos(c + d*x)**(5/2), x)
Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.54 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \, {\left (\frac {3 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \] Input:
integrate((a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")
Output:
4/3*(3*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sqrt(2)*a^(3/2) *sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(c os(d*x + c) + 1)^5)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d *x + c)/(cos(d*x + c) + 1) + 1)^(5/2))
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x, algorithm="giac")
Output:
Timed out
Time = 41.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,a\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (5\,\sin \left (c+d\,x\right )+2\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (3\,c+3\,d\,x\right )\right )}{3\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (3\,\cos \left (c+d\,x\right )+2\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (3\,c+3\,d\,x\right )+2\right )} \] Input:
int((a + a*cos(c + d*x))^(3/2)/cos(c + d*x)^(5/2),x)
Output:
(2*a*(a*(cos(c + d*x) + 1))^(1/2)*(5*sin(c + d*x) + 2*sin(2*c + 2*d*x) + 5 *sin(3*c + 3*d*x)))/(3*d*cos(c + d*x)^(1/2)*(3*cos(c + d*x) + 2*cos(2*c + 2*d*x) + cos(3*c + 3*d*x) + 2))
\[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\sqrt {a}\, a \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x +\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) \] Input:
int((a+a*cos(d*x+c))^(3/2)/cos(d*x+c)^(5/2),x)
Output:
sqrt(a)*a*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x)**3 ,x) + int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x))