Integrand size = 35, antiderivative size = 74 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {A b^2 \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:
A*b^2*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+b^2*C*(b *cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.59 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (A \text {arctanh}(\sin (c+d x))+C \sin (c+d x))}{d \cos ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/ 2),x]
Output:
((b*Cos[c + d*x])^(5/2)*(A*ArcTanh[Sin[c + d*x]] + C*Sin[c + d*x]))/(d*Cos [c + d*x]^(5/2))
Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2031, 3042, 3493, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec (c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \sec (c+d x)dx+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {A \text {arctanh}(\sin (c+d x))}{d}+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(7/2),x]
Output:
(b^2*Sqrt[b*Cos[c + d*x]]*((A*ArcTanh[Sin[c + d*x]])/d + (C*Sin[c + d*x])/ d))/Sqrt[Cos[c + d*x]]
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {b^{2} \left (2 A \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )-C \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}\) | \(56\) |
parts | \(-\frac {2 A \sqrt {b \cos \left (d x +c \right )}\, \operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) b^{2}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} C \sqrt {b \cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\) | \(77\) |
risch | \(-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\) | \(155\) |
Input:
int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RET URNVERBOSE)
Output:
-b^2/d*(2*A*arctanh(-csc(d*x+c)+cot(d*x+c))-C*sin(d*x+c))*(b*cos(d*x+c))^( 1/2)/cos(d*x+c)^(1/2)
Time = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.84 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left [\frac {A b^{\frac {5}{2}} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} C b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, -\frac {A \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right ) - \sqrt {b \cos \left (d x + c\right )} C b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algo rithm="fricas")
Output:
[1/2*(A*b^(5/2)*cos(d*x + c)*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))*C*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*co s(d*x + c)), -(A*sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c) - sqrt(b*cos(d*x + c))*C*b^2*sq rt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2),x)
Output:
Timed out
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, C b^{\frac {5}{2}} \sin \left (d x + c\right ) + {\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} A \sqrt {b}}{2 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algo rithm="maxima")
Output:
1/2*(2*C*b^(5/2)*sin(d*x + c) + (b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1))*A*sqrt(b))/d
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {{\left (A b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - A b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x, algo rithm="giac")
Output:
(A*b^2*log(tan(1/2*d*x + 1/2*c) + 1) - A*b^2*log(tan(1/2*d*x + 1/2*c) - 1) + 2*C*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1))*sqrt(b)/d
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(7/2),x)
Output:
int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(7/2), x)
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\sin \left (d x +c \right ) c \right )}{d} \] Input:
int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x)
Output:
(sqrt(b)*b**2*( - log(tan((c + d*x)/2) - 1)*a + log(tan((c + d*x)/2) + 1)* a + sin(c + d*x)*c))/d