\(\int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx\) [111]

Optimal result

Integrand size = 30, antiderivative size = 151 \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {c^3 \sec (e+f x) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

c^3*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)+4*c^3*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)-c^3*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 = 0.16 (sec) , antiderivative size = 151, 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, 84} \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=-\frac {c^3 \tan (e+f x) \sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[In]

Int[(c - c*Sec[e + f*x])^(5/2)/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(c^3*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (4*c^3*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]]) - (c^3*Sec[e + f*x]*Tan[e + f*
x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 {(c-c x)^2}{x (a+a 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 {c^2}{a}+\frac {c^2}{a x}-\frac {4 c^2}{a (1+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 {c^3 \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 {4 c^3 \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)}}-\frac {c^3 \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 = 0.54 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46 \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\frac {c^3 (\log (\cos (e+f x))+4 \log (1+\sec (e+f x))-\sec (e+f x)) \tan (e+f x)}{f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[(c - c*Sec[e + f*x])^(5/2)/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(c^3*(Log[Cos[e + f*x]] + 4*Log[1 + Sec[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 = 2.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \left (\sec \left (f x +e \right )-1\right )^{2} c^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+3 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+3 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\cos \left (f x +e \right )+1\right ) \cos \left (f x +e \right ) \cot \left (f x +e \right )}{f a \left (\cos \left (f x +e \right )-1\right )^{2}}\)	\(146\)
risch	\(-\frac {c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (-3 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )-2 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{3 i \left (f x +e \right )} f x -3 i {\mathrm e}^{i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+8 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )-2 i {\mathrm e}^{2 i \left (f x +e \right )}-3 i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) {\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{3 i \left (f x +e \right )} e +{\mathrm e}^{i \left (f x +e \right )} f x +{\mathrm e}^{2 i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}-3 i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+8 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+2 \,{\mathrm e}^{i \left (f x +e \right )} e +2 \,{\mathrm e}^{2 i \left (f x +e \right )} e +f x +2 e \right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}\)	\(372\)

[In]

int((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f/a*(-c*(sec(f*x+e)-1))^(1/2)*(sec(f*x+e)-1)^2*c^2*(a*(sec(f*x+e)+1))^(1/2)*(cos(f*x+e)*ln(2/(cos(f*x+e)+1))
+3*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)-1)+3*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)+cos(f*x+e)+1)/(cos(f*x+e)
-1)^2*cos(f*x+e)*cot(f*x+e)

Fricas [F]

\[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate((c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-2)]

Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]

[In]

int((c - c/cos(e + f*x))^(5/2)/(a + a/cos(e + f*x))^(1/2),x)

[Out]

int((c - c/cos(e + f*x))^(5/2)/(a + a/cos(e + f*x))^(1/2), x)