Integrand size = 43, antiderivative size = 102 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {C x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}}+\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]
A*sin(d*x+c)/b^2/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+C*x*cos(d*x+c)^(1 /2)/b^2/(b*cos(d*x+c))^(1/2)+B*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b^2/d/ (b*cos(d*x+c))^(1/2)
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) (C d x \cos (c+d x)+B \text {arctanh}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x))}{d (b \cos (c+d x))^{5/2}} \]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b* Cos[c + d*x])^(5/2),x]
(Cos[c + d*x]^(3/2)*(C*d*x*Cos[c + d*x] + B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x] + A*Sin[c + d*x]))/(d*(b*Cos[c + d*x])^(5/2))
Time = 0.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2031, 3042, 3500, 3042, 3214, 3042, 4257}
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.
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^2(c+d x)dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\int (B+C \cos (c+d x)) \sec (c+d x)dx+\frac {A \tan (c+d x)}{d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x)}{d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \sec (c+d x)dx+\frac {A \tan (c+d x)}{d}+C x\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {A \tan (c+d x)}{d}+C x\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {A \tan (c+d x)}{d}+\frac {B \text {arctanh}(\sin (c+d x))}{d}+C x\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*(C*x + (B*ArcTanh[Sin[c + d*x]])/d + (A*Tan[c + d*x])/ d))/(b^2*Sqrt[b*Cos[c + d*x]])
3.4.35.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 10.51 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {-2 B \cos \left (d x +c \right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+C \cos \left (d x +c \right ) \left (d x +c \right )+A \sin \left (d x +c \right )}{b^{2} d \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\) | \(73\) |
| parts | \(\frac {A \sin \left (d x +c \right )}{b^{2} d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}-\frac {2 B \left (\sqrt {\cos }\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}\, b^{2}}+\frac {C \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\) | \(108\) |
| risch | \(\frac {C x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {2 i \left (\sqrt {\cos }\left (d x +c \right )\right ) A}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(146\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(cos(d*x+c)*b)^(5/2), x,method=_RETURNVERBOSE)
1/b^2/d*(-2*B*cos(d*x+c)*arctanh(cot(d*x+c)-csc(d*x+c))+C*cos(d*x+c)*(d*x+ c)+A*sin(d*x+c))/(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(1/2)
Time = 0.35 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.11 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\left [-\frac {2 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} + C \sqrt {-b} \cos \left (d x + c\right )^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}, \frac {2 \, C \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right )^{2} + B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="fricas")
[-1/2*(2*B*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*s qrt(cos(d*x + c))))*cos(d*x + c)^2 + C*sqrt(-b)*cos(d*x + c)^2*log(2*b*cos (d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b) - 2*sqrt(b*cos(d*x + c))*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3* d*cos(d*x + c)^2), 1/2*(2*C*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c)^2 + B*sqrt(b)*cos(d*x + c)^2 *log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c) )*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c) )*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^2)]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.46 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}} + \frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {5}{2}}} + \frac {4 \, C \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {5}{2}}}}{2 \, d} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="maxima")
1/2*(4*A*sqrt(b)*sin(2*d*x + 2*c)/(b^3*cos(2*d*x + 2*c)^2 + b^3*sin(2*d*x + 2*c)^2 + 2*b^3*cos(2*d*x + 2*c) + b^3) + B*(log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*si n(d*x + c) + 1))/b^(5/2) + 4*C*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^( 5/2))/d
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(b*co s(d*x + c))^(5/2), x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]