3.4.2 \(\int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [302]

3.4.2.1 Optimal result
3.4.2.2 Mathematica [A] (verified)
3.4.2.3 Rubi [A] (verified)
3.4.2.4 Maple [A] (verified)
3.4.2.5 Fricas [A] (verification not implemented)
3.4.2.6 Sympy [F(-1)]
3.4.2.7 Maxima [A] (verification not implemented)
3.4.2.8 Giac [F]
3.4.2.9 Mupad [F(-1)]

3.4.2.1 Optimal result

Integrand size = 43, antiderivative size = 96 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b C x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b B \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \]

output
A*b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+b*C*x*(b*cos(d*x+c) 
)^(1/2)/cos(d*x+c)^(1/2)+b*B*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/co 
s(d*x+c)^(1/2)
 
3.4.2.2 Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{3/2} (C d x \cos (c+d x)+B \text {arctanh}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x))}{d \cos ^{\frac {5}{2}}(c+d x)} \]

input
Integrate[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(7/2),x]
 
output
((b*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*Cos[c + d*x]^(5/2))
 
3.4.2.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.53, 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.

\(\displaystyle \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec ^2(c+d x)dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\int (B+C \cos (c+d x)) \sec (c+d x)dx+\frac {A \tan (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (B \int \sec (c+d x)dx+\frac {A \tan (c+d x)}{d}+C x\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {A \tan (c+d x)}{d}+C x\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {A \tan (c+d x)}{d}+\frac {B \text {arctanh}(\sin (c+d x))}{d}+C x\right )}{\sqrt {\cos (c+d x)}}\)

input
Int[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(7/2),x]
 
output
(b*Sqrt[b*Cos[c + d*x]]*(C*x + (B*ArcTanh[Sin[c + d*x]])/d + (A*Tan[c + d* 
x])/d))/Sqrt[Cos[c + d*x]]
 

3.4.2.3.1 Defintions of rubi rules used

rule 2031
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]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3214
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]
 

rule 3500
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]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
3.4.2.4 Maple [A] (verified)

Time = 10.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.74

method result size
default \(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (-2 B \cos \left (d x +c \right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+C \cos \left (d x +c \right ) \left (d x +c \right )+A \sin \left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{\frac {3}{2}}}\) \(71\)
parts \(\frac {A b \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \cos \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 B \sqrt {\cos \left (d x +c \right ) b}\, b \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {C b \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\) \(102\)
risch \(\frac {b C x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {2 i b \sqrt {\cos \left (d x +c \right ) b}\, A}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {b \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\) \(138\)

input
int((cos(d*x+c)*b)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2), 
x,method=_RETURNVERBOSE)
 
output
b/d*(cos(d*x+c)*b)^(1/2)*(-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)^(3/2)
 
3.4.2.5 Fricas [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.29 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left [-\frac {2 \, B \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - C \sqrt {-b} b \cos \left (d x + c\right )^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} A b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}, \frac {2 \, C b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right )^{2} + B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}\right ] \]

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="fricas")
 
output
[-1/2*(2*B*sqrt(-b)*b*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b 
*sqrt(cos(d*x + c))))*cos(d*x + c)^2 - C*sqrt(-b)*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*b*sqrt(cos(d*x + c))*sin(d*x + c)) 
/(d*cos(d*x + c)^2), 1/2*(2*C*b^(3/2)*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*b^(3/2)*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*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)]
 
3.4.2.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

input
integrate((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c 
)**(7/2),x)
 
output
Timed out
 
3.4.2.7 Maxima [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.53 \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \, C b^{\frac {3}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + {\left (b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} B \sqrt {b} + \frac {4 \, A b^{\frac {3}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}}{2 \, d} \]

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="maxima")
 
output
1/2*(4*C*b^(3/2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + (b*log(cos(d*x 
+ c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b*log(cos(d*x + c)^2 + sin 
(d*x + c)^2 - 2*sin(d*x + c) + 1))*B*sqrt(b) + 4*A*b^(3/2)*sin(2*d*x + 2*c 
)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d
 
3.4.2.8 Giac [F]

\[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="giac")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(3/2)/c 
os(d*x + c)^(7/2), x)
 
3.4.2.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \]

input
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2),x)
 
output
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2), x)