\(\int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [528]

Optimal result

Integrand size = 35, antiderivative size = 144 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (2 A+3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

a^(3/2)*(2*A+3*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d 
*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+a^2*(2*A-B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/( 
a+a*sec(d*x+c))^(1/2)+a*B*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^( 
1/2)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a \sqrt {\cos (c+d x)} \sqrt {a (1+\sec (c+d x))} \left (2 A \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}-3 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}+\sqrt {1-\sec (c+d x)} (2 A+B \sec (c+d x))\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}} \] Input:

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

Output:

(a*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]*(2*A*ArcSin[Sqrt[1 - Sec[ 
c + d*x]]]*Sqrt[Sec[c + d*x]] - 3*B*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec[c 
+ d*x]] + Sqrt[1 - Sec[c + d*x]]*(2*A + B*Sec[c + d*x]))*Tan[(c + d*x)/2]) 
/(d*Sqrt[1 - Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 152, 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, 4506, 27, 3042, 4503, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4503

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a*B*Sqrt[Sec[c + d*x]]*Sqrt[a + a* 
Sec[c + d*x]]*Sin[c + d*x])/d + ((2*a^(3/2)*(2*A + 3*B)*ArcSinh[(Sqrt[a]*T 
an[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(2*A - B)*Sqrt[Sec[c + 
d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

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 4288

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

rule 4503

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

rule 4506

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(352\) vs. \(2(126)=252\).

Time = 2.44 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.45

method	result	size
default	\(-\frac {\left (\sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )\right )-3 B \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )\right )-3 B \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-B \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\cos \left (d x +c \right )}}\)	\(353\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.70 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {4 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

[1/4*(4*(2*A*a*cos(d*x + c) + B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
*sqrt(cos(d*x + c))*sin(d*x + c) + ((2*A + 3*B)*a*cos(d*x + c)^2 + (2*A + 
3*B)*a*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos 
(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x 
 + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*c 
os(d*x + c)^2 + d*cos(d*x + c)), 1/2*(2*(2*A*a*cos(d*x + c) + B*a)*sqrt((a 
*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((2*A + 
 3*B)*a*cos(d*x + c)^2 + (2*A + 3*B)*a*cos(d*x + c))*sqrt(-a)*arctan(2*sqr 
t(-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)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^2 + d*cos( 
d*x + c))]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1417 vs. \(2 (126) = 252\).

Time = 0.30 (sec) , antiderivative size = 1417, normalized size of antiderivative = 9.84 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2 
*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 
2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2 
*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt 
(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)* 
cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log 
(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d 
*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1/2* 
c))*A*sqrt(a) + (3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c 
)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) 
 - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*c 
os(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/ 
2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2* 
c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*si 
n(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2 
*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt 
(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2 
*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 
 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (126) = 252\).

Time = 0.50 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.42 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {\frac {4 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + {\left (2 \, A a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) - {\left (2 \, A a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) + \frac {4 \, {\left (3 \, \sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} B a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{\frac {7}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}}}{2 \, d} \] Input:

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

Output:

1/2*(4*sqrt(2)*A*a^2*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2 
*d*x + 1/2*c)^2 + a) + (2*A*a^(3/2)*sgn(cos(d*x + c)) + 3*B*a^(3/2)*sgn(co 
s(d*x + 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))^2 - a*(2*sqrt(2) + 3))) - (2*A*a^(3/2)*sgn(cos(d*x + c)) + 
3*B*a^(3/2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr 
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(3*sqrt(2)*(s 
qrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(5 
/2)*sgn(cos(d*x + c)) - sqrt(2)*B*a^(7/2)*sgn(cos(d*x + c)))/((sqrt(a)*tan 
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan 
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2,x 
)*b + int(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x),x)*a + in 
t(sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x),x)*b + int(sqrt(s 
ec(c + d*x) + 1)*sqrt(cos(c + d*x)),x)*a)