Integrand size = 43, antiderivative size = 111 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]
1/2*A*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+B*sin(d*x+c)/d/co s(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+1/2*(A+2*C)*arctanh(sin(d*x+c))*cos(d* x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(A+2 C) \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(A+2 B \cos (c+d x)) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*Sqrt [b*Cos[c + d*x]]),x]
((A + 2*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + (A + 2*B*Cos[c + d*x])*S in[c + d*x])/(2*d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]])
Time = 0.50 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {2032, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2032 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^3(c+d x)dx}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \int (2 B+(A+2 C) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \int \frac {2 B+(A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \sec (c+d x)dx+2 B \int \sec ^2(c+d x)dx\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 B \int 1d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \left ((A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 B \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \left (\frac {(A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 B \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*((A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((A + 2*C)*Arc Tanh[Sin[c + d*x]])/d + (2*B*Tan[c + d*x])/d)/2))/Sqrt[b*Cos[c + d*x]]
3.4.20.3.1 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[a*v]/Sqrt[b*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 10.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-A \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-4 C \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+2 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+A \sin \left (d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(130\) |
risch | \(-\frac {i \left (A \,{\mathrm e}^{2 i \left (d x +c \right )}-A -4 B \cos \left (d x +c \right )\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(145\) |
parts | \(\frac {A \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}-\frac {2 C \left (\sqrt {\cos }\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}}\) | \(155\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(cos(d*x+c)*b)^(1/2), x,method=_RETURNVERBOSE)
1/2/d*(A*cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)+1)-A*cos(d*x+c)^2*ln(-cot( d*x+c)+csc(d*x+c)-1)-4*C*cos(d*x+c)^2*arctanh(cot(d*x+c)-csc(d*x+c))+2*B*s in(d*x+c)*cos(d*x+c)+A*sin(d*x+c))/(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(3/2)
Time = 0.32 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.15 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {{\left (A + 2 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{3}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (1/2),x, algorithm="fricas")
[1/4*((A + 2*C)*sqrt(b)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*c os(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/c os(d*x + c)^3) + 2*(2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d* x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^3), -1/2*((A + 2*C)*sqrt(-b)*arcta n(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d *x + c)^3 - (2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c)) *sin(d*x + c))/(b*d*cos(d*x + c)^3)]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (95) = 190\).
Time = 0.53 (sec) , antiderivative size = 785, normalized size of antiderivative = 7.07 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (1/2),x, algorithm="maxima")
1/4*(2*C*(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*sin(d*x + c) + 1))/sqrt(b) + 8*B*sqrt( b)*sin(2*d*x + 2*c)/(b*cos(2*d*x + 2*c)^2 + b*sin(2*d*x + 2*c)^2 + 2*b*cos (2*d*x + 2*c) + b) - (4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*ar ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2 *d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2* cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*si n(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos (2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c ) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A/((2*(2*cos( 2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x +...
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^ (1/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c))*co s(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \]