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

Optimal result

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

[Out]

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*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/cos(d*x+c)
^(1/2)

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3034, 4103, 4100, 3886, 221} \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (2 A-B) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {a B \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Rule 221

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 3034

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(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] && IntegerQ[n])

Rule 3886

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

Rule 4100

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*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(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] /; Fr
eeQ[{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 4103

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {a B \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (2 A-B)+\frac {1}{2} a (2 A+3 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \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)}}+\frac {1}{2} \left (a (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx \\ & = \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)}}-\frac {\left (a (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a \sqrt {\cos (c+d x)} \sqrt {a (1+\sec (c+d x))} \left (2 A \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}-3 B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}+\sqrt {1-\sec (c+d x)} (2 A+B \sec (c+d x))\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

(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]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(307\) vs. \(2(126)=252\).

Time = 7.83 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.14

method	result	size
default	\(\frac {a \left (4 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-2 A \cos \left (d x +c \right ) \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-2 A \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 B \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-3 B \cos \left (d x +c \right ) \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 B \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\cos \left (d x +c \right )}}\)	\(308\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.70 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {4 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right ) + B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

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

[Out]

[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*c
os(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*cos(d*x + c)^2 + d*cos(d*x + c)), 1/2*(2*(2*A*a*cos(d*x + c) + B*a)*sqr
t((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*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)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]

Sympy [F(-1)]

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

[In]

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

[Out]

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 = 0.47 (sec) , antiderivative size = 1417, normalized size of antiderivative = 9.84 \[ \int \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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)*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)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*
sqrt(2)*sin(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*c
os(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*sqr
t(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*
x + 1/2*c) + 2*(2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 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) - 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) + 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) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) + 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) - 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) + 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) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 4*(sqrt(2)*a*cos(3/2*d*x + 3/2*c) - sqrt(2
)*a*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*B*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1))/d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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