\(\int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx\) [88]

Optimal result

Integrand size = 30, antiderivative size = 93 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=\frac {a c^2 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {a c \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3991, 3990, 3556} \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=\frac {a c^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f \sqrt {a \sec (e+f x)+a}} \]

[In]

Int[Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(a*c^2*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (a*c*Sqrt[c - c
*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]])

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3990

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

Rule 3991

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

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=\frac {a c^2 (\log (\cos (e+f x))+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\sec \left (f x +e \right )-1\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+\cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos \left (f x +e \right )+1\right ) \cot \left (f x +e \right )}{f \left (\cos \left (f x +e \right )-1\right )}\)	\(132\)
risch	\(\frac {c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, x}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (f x +e \right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {2 i c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, {\mathrm e}^{i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {i c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(428\)

[In]

int((c-c*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.76 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=\left [-\frac {2 \, c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \sqrt {-a c} {\left (c \cos \left (f x + e\right ) + c\right )} \log \left (\frac {a c \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + a c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \, {\left (f \cos \left (f x + e\right ) + f\right )}}, -\frac {c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - \sqrt {a c} {\left (c \cos \left (f x + e\right ) + c\right )} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )^{2} + a c}\right )}{f \cos \left (f x + e\right ) + f}\right ] \]

[In]

integrate((c-c*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) - sqrt
(-a*c)*(c*cos(f*x + e) + c)*log(1/2*(a*c*cos(f*x + e)^4 - (cos(f*x + e)^3 + cos(f*x + e))*sqrt(-a*c)*sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) + a*c)/cos(f*x + e)^2))/(f
*cos(f*x + e) + f), -(c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*
x + e) - sqrt(a*c)*(c*cos(f*x + e) + c)*arctan(sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f
*x + e) - c)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(a*c*cos(f*x + e)^2 + a*c)))/(f*cos(f*x + e) + f)]

Sympy [F]

\[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=\int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((c-c*sec(f*x+e))**(3/2)*(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*(-c*(sec(e + f*x) - 1))**(3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (85) = 170\).

Time = 0.40 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.61 \[ \int \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2} \, dx=-\frac {{\left ({\left (f x + e\right )} c \cos \left (2 \, f x + 2 \, e\right )^{2} + {\left (f x + e\right )} c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, {\left (f x + e\right )} c \cos \left (2 \, f x + 2 \, e\right ) + 2 \, c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (2 \, f x + 2 \, e\right ) + {\left (f x + e\right )} c - {\left (c \cos \left (2 \, f x + 2 \, e\right )^{2} + c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, {\left (c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} f} \]

[In]

integrate((c-c*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((c-c*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^(3/2),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^(3/2), x)