\(\int \frac {(a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx\) [97]

Optimal result

Integrand size = 30, antiderivative size = 104 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {a^2 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 a^2 \log (1-\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3997, 78} \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^2 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 3997

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

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.62 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {a (\log (\cos (e+f x))+2 \log (1-\sec (e+f x))) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{f \sqrt {c-c \sec (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )\right ) \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\)	\(112\)
risch	\(\frac {a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}}-\frac {2 a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}-\frac {4 i a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}+\frac {i a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\)	\(390\)

[In]

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

[Out]

1/f*a*(a*(sec(f*x+e)+1))^(1/2)*(ln(2/(cos(f*x+e)+1))+ln(-cot(f*x+e)+csc(f*x+e)+1)-4*ln(-cot(f*x+e)+csc(f*x+e))
+ln(-cot(f*x+e)+csc(f*x+e)-1))/(-c*(sec(f*x+e)-1))^(1/2)*(cot(f*x+e)-csc(f*x+e))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=-\frac {{\left ({\left (f x + e\right )} a + a \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, a \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right )\right )} \sqrt {a}}{\sqrt {c} f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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