3.177 \(\int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=226 \[ -\frac {c g (3-4 p) \sin (e+f x) (g \sec (e+f x))^{p-1} \, _2F_1\left (\frac {1}{2},\frac {1-p}{2};\frac {3-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {c (5-4 p) \sin (e+f x) (g \sec (e+f x))^p \, _2F_1\left (\frac {1}{2},-\frac {p}{2};\frac {2-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}-\frac {c (5-4 p) \tan (e+f x) (g \sec (e+f x))^p}{3 a^2 f (\sec (e+f x)+1)}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-1/3*c*g*(3-4*p)*hypergeom([1/2, 1/2-1/2*p],[3/2-1/2*p],cos(f*x+e)^2)*(g*sec(f*x+e))^(-1+p)*sin(f*x+e)/a^2/f/(
sin(f*x+e)^2)^(1/2)+1/3*c*(5-4*p)*hypergeom([1/2, -1/2*p],[1-1/2*p],cos(f*x+e)^2)*(g*sec(f*x+e))^p*sin(f*x+e)/
a^2/f/(sin(f*x+e)^2)^(1/2)-1/3*c*(5-4*p)*(g*sec(f*x+e))^p*tan(f*x+e)/a^2/f/(1+sec(f*x+e))-2/3*c*(g*sec(f*x+e))
^p*tan(f*x+e)/f/(a+a*sec(f*x+e))^2

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Rubi [A]  time = 0.41, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4020, 3787, 3772, 2643} \[ -\frac {c g (3-4 p) \sin (e+f x) (g \sec (e+f x))^{p-1} \, _2F_1\left (\frac {1}{2},\frac {1-p}{2};\frac {3-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {c (5-4 p) \sin (e+f x) (g \sec (e+f x))^p \, _2F_1\left (\frac {1}{2},-\frac {p}{2};\frac {2-p}{2};\cos ^2(e+f x)\right )}{3 a^2 f \sqrt {\sin ^2(e+f x)}}-\frac {c (5-4 p) \tan (e+f x) (g \sec (e+f x))^p}{3 a^2 f (\sec (e+f x)+1)}-\frac {2 c \tan (e+f x) (g \sec (e+f x))^p}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*g*(3 - 4*p)*Hypergeometric2F1[1/2, (1 - p)/2, (3 - p)/2, Cos[e + f*x]^2]*(g*Sec[e + f*x])^(-1 + p)*Sin[e +
f*x])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) + (c*(5 - 4*p)*Hypergeometric2F1[1/2, -p/2, (2 - p)/2, Cos[e + f*x]^2]*(
g*Sec[e + f*x])^p*Sin[e + f*x])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) - (c*(5 - 4*p)*(g*Sec[e + f*x])^p*Tan[e + f*x])
/(3*a^2*f*(1 + Sec[e + f*x])) - (2*c*(g*Sec[e + f*x])^p*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx &=-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\int \frac {(g \sec (e+f x))^p (a c (3-2 p)-2 a c (1-p) \sec (e+f x))}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac {c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\int (g \sec (e+f x))^p \left (a^2 c (3-4 p) (1-p)+a^2 c (5-4 p) p \sec (e+f x)\right ) \, dx}{3 a^4}\\ &=-\frac {c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {(c (3-4 p) (1-p)) \int (g \sec (e+f x))^p \, dx}{3 a^2}+\frac {(c (5-4 p) p) \int (g \sec (e+f x))^{1+p} \, dx}{3 a^2 g}\\ &=-\frac {c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (c (3-4 p) (1-p) \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p\right ) \int \left (\frac {\cos (e+f x)}{g}\right )^{-p} \, dx}{3 a^2}+\frac {\left (c (5-4 p) p \left (\frac {\cos (e+f x)}{g}\right )^p (g \sec (e+f x))^p\right ) \int \left (\frac {\cos (e+f x)}{g}\right )^{-1-p} \, dx}{3 a^2 g}\\ &=-\frac {c (3-4 p) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-p}{2};\frac {3-p}{2};\cos ^2(e+f x)\right ) (g \sec (e+f x))^p \sin (e+f x)}{3 a^2 f \sqrt {\sin ^2(e+f x)}}+\frac {c (5-4 p) \, _2F_1\left (\frac {1}{2},-\frac {p}{2};\frac {2-p}{2};\cos ^2(e+f x)\right ) (g \sec (e+f x))^p \sin (e+f x)}{3 a^2 f \sqrt {\sin ^2(e+f x)}}-\frac {c (5-4 p) (g \sec (e+f x))^p \tan (e+f x)}{3 a^2 f (1+\sec (e+f x))}-\frac {2 c (g \sec (e+f x))^p \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end {align*}

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Mathematica [F]  time = 13.33, size = 0, normalized size = 0.00 \[ \int \frac {(g \sec (e+f x))^p (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*
pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(
-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to ch
eck sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(
-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*
pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to divide, perhaps
due to rounding error%%%{-1,[0,1,4,0]%%%}+%%%{1,[0,1,0,0]%%%} / %%%{4,[0,0,0,2]%%%} Error: Bad Argument Value

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maple [F]  time = 2.52, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \sec \left (f x +e \right )\right )^{p} \left (c -c \sec \left (f x +e \right )\right )}{\left (a +a \sec \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^2,x)

[Out]

int((g*sec(f*x+e))^p*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (c \sec \left (f x + e\right ) - c\right )} \left (g \sec \left (f x + e\right )\right )^{p}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((c*sec(f*x + e) - c)*(g*sec(f*x + e))^p/(a*sec(f*x + e) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )\,{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^p}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {c \left (\int \left (- \frac {\left (g \sec {\left (e + f x \right )}\right )^{p}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\left (g \sec {\left (e + f x \right )}\right )^{p} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(Integral(-(g*sec(e + f*x))**p/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral((g*sec(e + f*x))**p*se
c(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2

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