\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [1168]

Optimal result

Integrand size = 37, antiderivative size = 266 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(163 A+19 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(299 A+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {5 (19 A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/32*(163*A+19*C)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/ 
(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(5/2)/ 
d-1/4*(A+C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(17* 
A+C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/48*(299*A+27 
*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+5/48*(19*A+3* 
C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 3.38 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {6 (163 A+19 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )-(379 A+27 C+(479 A+39 C) \cos (c+d x)+80 A \cos (2 (c+d x))-8 A \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{48 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]) 
^(5/2),x]
 

Output:

(6*(163*A + 19*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 - (379*A + 
27*C + (479*A + 39*C)*Cos[c + d*x] + 80*A*Cos[2*(c + d*x)] - 8*A*Cos[3*(c 
+ d*x)])*Tan[(c + d*x)/2])/(48*a^2*d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x]) 
*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.71 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 4753, 3042, 4573, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

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.

\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^{3/2} \left (A+C \sec (c+d x)^2\right )}{(a \sec (c+d x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 4753

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+A}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4573

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int -\frac {a (11 A+3 C)-2 a (3 A-C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A+3 C)-2 a (3 A-C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (11 A+3 C)-2 a (3 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^3 (163 A+19 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^3 (163 A+19 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^3 (163 A+19 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (299 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{5/2} (163 A+19 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\right )\)

Input:

Int[(Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2) 
,x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A + C)*Sin[c + d*x])/(d*Sqrt 
[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(17*A + C)*Sin[c + d 
*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((10*a^2*(19*A + 
3*C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((- 
3*Sqrt[2]*a^(5/2)*(163*A + 19*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c 
 + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^3*(299*A + 27*C)*Sq 
rt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2 
))/(8*a^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4295

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f))   Subst[Int[1/(2*b - d*x^2), x], 
x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; 
 FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4501

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

rule 4508

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b 
- a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] - Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs 
c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ 
e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B 
, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]
 

rule 4510

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

rule 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m 
+ 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - 
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [A] (verified)

Time = 2.90 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.19

method	result	size
default	\(-\frac {\left (\left (489 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+1467 \cos \left (d x +c \right )+489\right ) A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-57 \cos \left (d x +c \right )^{2}-114 \cos \left (d x +c \right )-57\right ) \sqrt {2}\, C \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \left (39 \cos \left (d x +c \right )+27\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\left (-64 \cos \left (d x +c \right )^{3}+320 \cos \left (d x +c \right )^{2}+1006 \cos \left (d x +c \right )+598\right ) \sin \left (d x +c \right ) A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{96 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{3} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(316\)

Input:

int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-1/96/d*((489*cos(d*x+c)^3+1467*cos(d*x+c)^2+1467*cos(d*x+c)+489)*A*(-2/(c 
os(d*x+c)+1))^(1/2)*(-1/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-1/(cos( 
d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))+(-57*cos(d*x+c)^2-114*cos(d*x+c 
)-57)*2^(1/2)*C*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+ 
cot(d*x+c)))+sin(d*x+c)*(39*cos(d*x+c)+27)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^( 
1/2)+(-64*cos(d*x+c)^3+320*cos(d*x+c)^2+1006*cos(d*x+c)+598)*sin(d*x+c)*A* 
(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos( 
d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)/a^3/(-1/(cos(d*x+c)+1))^(1/2)
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 538, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

[1/192*(3*sqrt(2)*((163*A + 19*C)*cos(d*x + c)^3 + 3*(163*A + 19*C)*cos(d* 
x + c)^2 + 3*(163*A + 19*C)*cos(d*x + c) + 163*A + 19*C)*sqrt(a)*log(-(a*c 
os(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 
+ 2*cos(d*x + c) + 1)) + 4*(32*A*cos(d*x + c)^3 - 160*A*cos(d*x + c)^2 - ( 
503*A + 39*C)*cos(d*x + c) - 299*A - 27*C)*sqrt((a*cos(d*x + c) + a)/cos(d 
*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d* 
cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/96*(3*sqrt(2)*((163*A + 
 19*C)*cos(d*x + c)^3 + 3*(163*A + 19*C)*cos(d*x + c)^2 + 3*(163*A + 19*C) 
*cos(d*x + c) + 163*A + 19*C)*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((a 
*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d* 
x + c) + a)) - 2*(32*A*cos(d*x + c)^3 - 160*A*cos(d*x + c)^2 - (503*A + 39 
*C)*cos(d*x + c) - 299*A - 27*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s 
qrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + 
c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151873 vs. \(2 (225) = 450\).

Time = 2.87 (sec) , antiderivative size = 151873, normalized size of antiderivative = 570.95 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

1/96*((32*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*si 
n(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3* 
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(11/3*arctan2(sin 
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 41472*(cos(3*d*x + 3*c)^2*si 
n(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d 
*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co 
s(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
3/2*c)))^4 + 8192*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3 
*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)* 
sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arct 
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 288*sin(3/2*d*x + 3/2 
*c)^5 + 32*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*s 
in(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3 
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(11/3*arctan2(si 
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 41472*(cos(3*d*x + 3*c)^2*s 
in(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3* 
d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), c 
os(3/2*d*x + 3/2*c))))*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
 3/2*c)))^4 + 8192*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 
3*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)...
 

Giac [A] (verification not implemented)

Time = 175.32 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} + C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (23 \, A a^{5} + 7 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {4 \, \sqrt {2} {\left (167 \, A a^{5} + 15 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} {\left (155 \, A a^{5} + 11 \, C a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {3 \, \sqrt {2} {\left (163 \, A + 19 \, C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{96 \, d} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

1/96*(((3*(2*sqrt(2)*(A*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^6*sgn(cos(d 
*x + c))) - sqrt(2)*(23*A*a^5 + 7*C*a^5)/(a^6*sgn(cos(d*x + c))))*tan(1/2* 
d*x + 1/2*c)^2 - 4*sqrt(2)*(167*A*a^5 + 15*C*a^5)/(a^6*sgn(cos(d*x + c)))) 
*tan(1/2*d*x + 1/2*c)^2 - 3*sqrt(2)*(155*A*a^5 + 11*C*a^5)/(a^6*sgn(cos(d* 
x + c))))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 3*sq 
rt(2)*(163*A + 19*C)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/ 
2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((cos(c + d*x)^(3/2)*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(5/2) 
,x)
 

Output:

int((cos(c + d*x)^(3/2)*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(5/2) 
, x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)*sec( 
c + d*x)**2)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x) 
*c + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x))/(sec(c + 
 d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3