\(\int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\) [124]

Optimal result

Integrand size = 30, antiderivative size = 220 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\frac {c^4 \log (\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^4 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 220, 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, 90} \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\frac {4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1) \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1)^2 \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^4 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {c^4 \tan (e+f x) \log (\cos (e+f x))}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[In]

Int[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(c^4*Log[Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (2*c^4*Log[1
+ Sec[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (4*c^4*Tan[e + f*x])
/(a^2*f*(1 + Sec[e + f*x])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (4*c^4*Tan[e + f*x])/(a^2*f*
(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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)^3}{x (a+a x)^3} \, 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^3}{a^3 x}-\frac {8 c^3}{a^3 (1+x)^3}+\frac {4 c^3}{a^3 (1+x)^2}-\frac {2 c^3}{a^3 (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^4 \log (\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^4 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.39 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^4 \left (-\log (\cos (e+f x))-2 \log (1+\sec (e+f x))-\frac {4 \sec (e+f x)}{(1+\sec (e+f x))^2}\right ) \tan (e+f x)}{a^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

-((c^4*(-Log[Cos[e + f*x]] - 2*Log[1 + Sec[e + f*x]] - (4*Sec[e + f*x])/(1 + Sec[e + f*x])^2)*Tan[e + f*x])/(a
^2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]]))

Maple [A] (warning: unable to verify)

Time = 2.36 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{4} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {7}{2}} \sin \left (f x +e \right )^{7} \left (\left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )\right )}{2 f \,a^{3} \left (1-\cos \left (f x +e \right )\right )^{7}}\)	\(216\)
risch	\(-\frac {c^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (-4 i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) {\mathrm e}^{3 i \left (f x +e \right )}+8 i {\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )} f x +2 \,{\mathrm e}^{4 i \left (f x +e \right )} e -4 i {\mathrm e}^{i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+16 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}+4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +8 \,{\mathrm e}^{3 i \left (f x +e \right )} e +16 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+4 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{4 i \left (f x +e \right )}+6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x +12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +24 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 i {\mathrm e}^{3 i \left (f x +e \right )}+4 \,{\mathrm e}^{i \left (f x +e \right )} f x -i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e -i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )-6 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+4 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+f x +2 e \right )}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(450\)

[In]

int((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*2^(1/2)/a^3*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^4*(c*(1-co
s(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)*csc(f*x+e)^2)^(7/2)/(1-cos(f*x+e))^7*sin(f*x+e)^7*((1-cos(f*x+e)
)^4*csc(f*x+e)^4+ln(-cot(f*x+e)+csc(f*x+e)+1)+ln((1-cos(f*x+e))^2*csc(f*x+e)^2+1)+ln(-cot(f*x+e)+csc(f*x+e)-1)
)

Fricas [F]

\[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.42 \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {{\left ({\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{2} {\left | c \right |} + 2 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{2} c {\left | c \right |}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{5} f} \]

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^2*abs(c) + 2*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^2*c*a
bs(c))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/(a^5*f)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((c - c/cos(e + f*x))^(7/2)/(a + a/cos(e + f*x))^(5/2),x)

[Out]

int((c - c/cos(e + f*x))^(7/2)/(a + a/cos(e + f*x))^(5/2), x)