\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx\) [551]

Optimal result

Integrand size = 35, antiderivative size = 127 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(3 A+B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \] Output:

1/4*(3*A+B)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*s 
ec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d-1/2* 
(A-B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(325\) vs. \(2(127)=254\).

Time = 0.74 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.56 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-\sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x)-\sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x)-2 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+2 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+2 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))+2 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x))\right ) \sin (c+d x)}{4 a d \sqrt {-1+\cos (c+d x)} (1+\cos (c+d x)) \sqrt {\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

((-3*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] 
 - Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] - 
 3*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*S 
ec[c + d*x] - Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c 
 + d*x]]]*Sec[c + d*x] - 2*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 2 
*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 2*B*ArcSin[Sqrt[1 - Sec[c + 
 d*x]]]*(1 + Sec[c + d*x]) + 2*B*ArcSin[Sqrt[Sec[c + d*x]]]*(1 + Sec[c + d 
*x]))*Sin[c + d*x])/(4*a*d*Sqrt[-1 + Cos[c + d*x]]*(1 + Cos[c + d*x])*Sqrt 
[Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3434, 3042, 4500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {(3 A+B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{2 a d}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(3 A+B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((3*A + B)*ArcTanh[(Sqrt[a]*Sqrt[Se 
c[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]* 
a^(3/2)*d) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + 
 d*x])^(3/2)))
 

Defintions of rubi rules used

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 3434

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])
 

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 4500

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[(a*A*m + b*B*(m + 1))/(a^2*(2*m + 1))   Int[(a + b*Csc[e + 
 f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n} 
, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && LeQ[ 
m, -1]
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.54

method	result	size
default	\(-\frac {\left (\left (-3 \cos \left (d x +c \right )-3\right ) A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\left (-\cos \left (d x +c \right )-1\right ) B \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} a^{2} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(196\)

Input:

int((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/4/d*((-3*cos(d*x+c)-3)*A*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(- 
1/(1+cos(d*x+c)))^(1/2))+A*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+(-cos(d*x+ 
c)-1)*B*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1 
/2))-B*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*2^(1/2)*(a*(1+sec(d*x+c)))^(1 
/2)*cos(d*x+c)^(1/2)/(1+cos(d*x+c))^2/a^2/(-1/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 384, normalized size of antiderivative = 3.02 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right ) + 3 \, A + B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (A - B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {\sqrt {2} {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right ) + 3 \, A + B\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + a\right )}}\right ) + 2 \, {\left (A - B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="fricas")
 

Output:

[1/8*(sqrt(2)*((3*A + B)*cos(d*x + c)^2 + 2*(3*A + B)*cos(d*x + c) + 3*A + 
 B)*sqrt(a)*log(-(a*cos(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*(A - B)*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c) 
^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/4*(sqrt(2)*((3*A + B)*cos(d*x + c)^ 
2 + 2*(3*A + B)*cos(d*x + c) + 3*A + B)*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*(A - B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x 
+ c) + a^2*d)]
 

Sympy [F]

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

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2166 vs. \(2 (104) = 208\).

Time = 0.29 (sec) , antiderivative size = 2166, normalized size of antiderivative = 17.06 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="maxima")
 

Output:

1/4*((3*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d 
*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2 
*sin(1/2*d*x + 1/2*c) + 1))*cos(2*d*x + 2*c)^2 + 12*(log(cos(1/2*d*x + 1/2 
*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2 
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*co 
s(d*x + c)^2 + 3*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* 
sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2 
*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(2*d*x + 2*c)^2 + 12*(log(cos(1/2* 
d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - lo 
g(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) 
 + 1))*sin(d*x + c)^2 + 2*(6*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1 
/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1 
/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c) + 3*log(cos( 
1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) 
- 3*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 
1/2*c) + 1) - 2*sin(3/2*d*x + 3/2*c) + 2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 
 2*c) + 4*(3*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1 
/2*d*x + 1/2*c) + 1) - 3*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 - 2*sin(1/2*d*x + 1/2*c) + 1) + 2*sin(1/2*d*x + 1/2*c))*cos(d*x + c) + 
4*(3*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d...
 

Giac [A] (verification not implemented)

Time = 167.93 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left (3 \, \sqrt {2} A + \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}}}{4 \, d} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="giac")
 

Output:

-1/4*((3*sqrt(2)*A + sqrt(2)*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(3/2)*sgn(cos(d*x + c))) + (sqrt(2)* 
A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)))*sqrt(a*tan(1/2*d*x 
+ 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3)/d
 

Mupad [F(-1)]

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

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))^(3/2)),x 
)
 

Output:

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))^(3/2)), 
x)
 

Reduce [F]

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

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

Output:

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