Integrand size = 43, antiderivative size = 164 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]
1/3*A*sin(d*x+c)/b^2/d/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+1/2*B*sin(d*x +c)/b^2/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/3*(2*A+3*C)*sin(d*x+c)/b ^2/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+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 = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \left (3 B \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+(4 A+3 C+3 B \cos (c+d x)+(2 A+3 C) \cos (2 (c+d x))) \tan (c+d x)\right )}{6 d (b \cos (c+d x))^{5/2}} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(b*C os[c + d*x])^(5/2)),x]
(Sqrt[Cos[c + d*x]]*(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + (4*A + 3*C + 3*B*Cos[c + d*x] + (2*A + 3*C)*Cos[2*(c + d*x)])*Tan[c + d*x]))/(6*d*(b *Cos[c + d*x])^(5/2))
Time = 0.62 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2032, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2032 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^4(c+d x)dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int (3 B+(2 A+3 C) \cos (c+d x)) \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int \frac {3 B+(2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left ((2 A+3 C) \int \sec ^2(c+d x)dx+3 B \int \sec ^3(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left ((2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {(2 A+3 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (3 B \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 A+3 C) \tan (c+d x)}{d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (\frac {(2 A+3 C) \tan (c+d x)}{d}+3 B \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*((A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((2*A + 3*C) *Tan[c + d*x])/d + 3*B*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3))/(b^2*Sqrt[b*Cos[c + d*x]])
3.4.37.3.1 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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.53 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.87
| method | result | size |
| default | \(\frac {3 B \left (\cos ^{3}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-3 B \left (\cos ^{3}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+4 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+6 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )}{6 b^{2} d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(142\) |
| parts | \(\frac {A \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sin \left (d x +c \right )}{3 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {B \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}+\frac {C \sin \left (d x +c \right )}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\) | \(166\) |
| risch | \(-\frac {i \left (3 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 B +\left (-16 A -18 C \right ) \cos \left (d x +c \right )+i \left (-8 A -6 C \right ) \sin \left (d x +c \right )\right )}{6 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(180\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(cos(d*x+c)*b)^(5/2), x,method=_RETURNVERBOSE)
1/6/b^2/d*(3*B*cos(d*x+c)^3*ln(-cot(d*x+c)+csc(d*x+c)+1)-3*B*cos(d*x+c)^3* ln(-cot(d*x+c)+csc(d*x+c)-1)+4*A*sin(d*x+c)*cos(d*x+c)^2+6*C*cos(d*x+c)^2* sin(d*x+c)+3*B*sin(d*x+c)*cos(d*x+c)+2*A*sin(d*x+c))/(cos(d*x+c)*b)^(1/2)/ cos(d*x+c)^(5/2)
Time = 0.30 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.65 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\left [\frac {3 \, B \sqrt {b} \cos \left (d x + c\right )^{4} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{3} d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{4} - {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{3} d \cos \left (d x + c\right )^{4}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="fricas")
[1/12*(3*B*sqrt(b)*cos(d*x + c)^4*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*(2*(2*A + 3*C)*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sqrt (b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^4), -1/6*(3*B*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sq rt(cos(d*x + c))))*cos(d*x + c)^4 - (2*(2*A + 3*C)*cos(d*x + c)^2 + 3*B*co s(d*x + c) + 2*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b ^3*d*cos(d*x + c)^4)]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (140) = 280\).
Time = 0.52 (sec) , antiderivative size = 1098, normalized size of antiderivative = 6.70 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="maxima")
1/12*(24*C*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) + 16*((3*cos(2*d*x + 2*c) + 1)* sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 3*cos(6*d *x + 6*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*A/((b^2* cos(6*d*x + 6*c)^2 + 9*b^2*cos(4*d*x + 4*c)^2 + 9*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(6*d*x + 6*c)^2 + 9*b^2*sin(4*d*x + 4*c)^2 + 18*b^2*sin(4*d*x + 4* c)*sin(2*d*x + 2*c) + 9*b^2*sin(2*d*x + 2*c)^2 + 6*b^2*cos(2*d*x + 2*c) + b^2 + 2*(3*b^2*cos(4*d*x + 4*c) + 3*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) + 6*(3*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 6*(b^2*sin(4* d*x + 4*c) + b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*sqrt(b)) - 3*(4*(sin( 4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(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*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*arctan 2(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 + ...
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^ (5/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c))^(5/2)* cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]