Integrand size = 35, antiderivative size = 162 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 (A-3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \] Output:
2^(1/2)*(A-B)*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)/a^(1/2)/d-2/3*(A-3*B )*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+2/3*A*cos(d*x+c)^(1 /2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (2 (-A+3 B+A \cos (c+d x)) \sqrt {1-\sec (c+d x)}-3 \sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) \sin (c+d x)}{3 d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x ]],x]
Output:
((2*(-A + 3*B + A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] - 3*Sqrt[2]*(A - B) *ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d*x]])*Sin[c + d*x])/(3*d*Sqrt[-1 + Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x] )])
Time = 0.92 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3434, 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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 3434 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
\(\Big \downarrow \) 4510 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {a (A-3 B)-2 a A \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-3 B)-2 a A \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-3 B)-2 a A \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
\(\Big \downarrow \) 4501 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a (A-B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
\(\Big \downarrow \) 4295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a (A-B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}\right )\) |
Input:
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/Sqrt[a + a*Sec[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-3*Sqrt[2]*Sqrt[a]*(A - B)*ArcTanh[( Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]] )])/d + (2*a*(A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[ c + d*x]]))/(3*a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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]
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]
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]
Time = 1.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {\left (\left (-3 \cos \left (d x +c \right )-3\right ) A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (3 \cos \left (d x +c \right )+3\right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-2+2 \cos \left (d x +c \right )\right ) \sin \left (d x +c \right ) A +6 B \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{3 d a \left (1+\cos \left (d x +c \right )\right )}\) | \(189\) |
Input:
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RET URNVERBOSE)
Output:
1/3/d*((-3*cos(d*x+c)-3)*A*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(- csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+(3*cos(d*x+c)+3)*B*(-2/( 1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+co s(d*x+c)))^(1/2))+(-2+2*cos(d*x+c))*sin(d*x+c)*A+6*B*sin(d*x+c))*cos(d*x+c )^(1/2)*(a*(1+sec(d*x+c)))^(1/2)/a/(1+cos(d*x+c))
Time = 0.10 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.13 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right ) - 2 \, {\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo rithm="fricas")
Output:
[1/6*(4*(A*cos(d*x + c) - A + 3*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) *sqrt(cos(d*x + c))*sin(d*x + c) - 3*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos( d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d), -1/3* (3*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*sqrt(-1/a)*arctan(1/2*sqrt (2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))* sin(d*x + c)/(cos(d*x + c) + 1)) - 2*(A*cos(d*x + c) - A + 3*B)*sqrt((a*co s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d* x + c) + a*d)]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(1/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (135) = 270\).
Time = 0.27 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.95 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo rithm="maxima")
Output:
-1/6*((3*sqrt(2)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c )))*sin(3/2*d*x + 3/2*c) - 3*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2( sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*log(cos(1/3*arcta n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/ 2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 3*sqrt(2)*log(cos(1/3*arctan2(sin(3 /2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)* sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A/sqrt(a) + 3*(sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + si n(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - sqrt(2)*log(cos(1/4*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c)))^2 - 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) ) + 1) - 4*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B /sqrt(a))/d
Time = 171.56 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, {\left (\frac {\sqrt {2} {\left (2 \, A a - 3 \, B a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {3 \, \sqrt {2} B a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{3 \, d} \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo rithm="giac")
Output:
-1/3*(3*sqrt(2)*(A - B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan (1/2*d*x + 1/2*c)^2 + a)))/(sqrt(a)*sgn(cos(d*x + c))) + 2*(sqrt(2)*(2*A*a - 3*B*a)*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 3*sqrt(2)*B*a/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))/d
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2),x )
Output:
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x)))/(a + a/cos(c + d*x))^(1/2), x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)*sec( c + d*x))/(sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x))/(sec(c + d*x) + 1),x)*a))/a