\(\int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx\) [45]

Optimal result

Integrand size = 26, antiderivative size = 66 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\frac {2 \sqrt {a} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}-\frac {2 a c \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}} \]

[Out]

2*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)/f-2*a*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3989, 3972, 327, 209} \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\frac {2 \sqrt {a} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f}-\frac {2 a c \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a}} \]

[In]

Int[Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x]),x]

[Out]

(2*Sqrt[a]*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a*c*Tan[e + f*x])/(f*Sqrt[a + a*S
ec[e + f*x]])

Rule 209

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 3972

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[-2*(a^(m/2 +
n + 1/2)/d), Subst[Int[x^m*((2 + a*x^2)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 3989

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[((-a)*c)^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps \begin{align*} \text {integral}& = -\left ((a c) \int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx\right ) \\ & = \frac {\left (2 a^2 c\right ) \text {Subst}\left (\int \frac {x^2}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f} \\ & = -\frac {2 a c \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}}-\frac {(2 a c) \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f} \\ & = \frac {2 \sqrt {a} c \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}-\frac {2 a c \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.38 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\frac {2 c \sqrt {a (1+\sec (e+f x))} \left (\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {c}}\right )-\sqrt {c-c \sec (e+f x)}\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{f \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x]),x]

[Out]

(2*c*Sqrt[a*(1 + Sec[e + f*x])]*(Sqrt[c]*ArcTanh[Sqrt[c - c*Sec[e + f*x]]/Sqrt[c]] - Sqrt[c - c*Sec[e + f*x]])
*Tan[(e + f*x)/2])/(f*Sqrt[c - c*Sec[e + f*x]])

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41

method	result	size
default	\(\frac {2 c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f}\)	\(93\)
parts	\(\frac {2 c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )}{f}+\frac {2 c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f}\)	\(111\)

[In]

int((c-c*sec(f*x+e))*(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*c/f*(a*(sec(f*x+e)+1))^(1/2)*(arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*
x+e)/(cos(f*x+e)+1))^(1/2)+cot(f*x+e)-csc(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 3.55 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\left [\frac {{\left (c \cos \left (f x + e\right ) + c\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{f \cos \left (f x + e\right ) + f}, -\frac {2 \, {\left ({\left (c \cos \left (f x + e\right ) + c\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{f \cos \left (f x + e\right ) + f}\right ] \]

[In]

integrate((c-c*sec(f*x+e))*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[((c*cos(f*x + e) + c)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c
os(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e
))*sin(f*x + e))/(f*cos(f*x + e) + f), -2*((c*cos(f*x + e) + c)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f
*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos
(f*x + e) + f)]

Sympy [F]

\[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=- c \left (\int \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}\, dx + \int \left (- \sqrt {a \sec {\left (e + f x \right )} + a}\right )\, dx\right ) \]

[In]

integrate((c-c*sec(f*x+e))*(a+a*sec(f*x+e))**(1/2),x)

[Out]

-c*(Integral(sqrt(a*sec(e + f*x) + a)*sec(e + f*x), x) + Integral(-sqrt(a*sec(e + f*x) + a), x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (58) = 116\).

Time = 0.37 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.23 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\frac {\sqrt {a} c \arctan \left ({\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + \sin \left (f x + e\right ), {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + \cos \left (f x + e\right )\right )}{f} \]

[In]

integrate((c-c*sec(f*x+e))*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

sqrt(a)*c*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + sin(f*x + e), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x +
2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + cos(f*x + e))/f

Giac [F]

\[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x)) \, dx=\int { -\sqrt {a \sec \left (f x + e\right ) + a} {\left (c \sec \left (f x + e\right ) - c\right )} \,d x } \]

[In]

integrate((c-c*sec(f*x+e))*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x)),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x)), x)