\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [227]

Optimal result

Integrand size = 35, antiderivative size = 76 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \sqrt {a} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a A \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {4100, 3886, 221} \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}+\frac {2 \sqrt {a} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 a \left (A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)-B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(161\) vs. \(2(66)=132\).

Time = 6.52 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.13

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (B \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}-B \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 A \cot \left (d x +c \right )-2 A \csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}\)	\(162\)
parts	\(-\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (\arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \cos \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(185\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (66) = 132\).

Time = 0.34 (sec) , antiderivative size = 307, normalized size of antiderivative = 4.04 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{d \cos \left (d x + c\right ) + d}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right )}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (66) = 132\).

Time = 0.49 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.45 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {4 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B \sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*(4*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + B*sqrt(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) - 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) + 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) - 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)))/d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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