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

Optimal. Leaf size=70 \[ -\frac {c \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {7 c \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {2 c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-c*arctanh(sin(f*x+e))/a^2/f+7/3*c*tan(f*x+e)/a^2/f/(1+sec(f*x+e))-2/3*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^2

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Rubi [A]  time = 0.16, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4008, 3998, 3770, 3794} \[ -\frac {c \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {7 c \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac {2 c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*ArcTanh[Sin[e + f*x]])/(a^2*f)) + (7*c*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - (2*c*Tan[e + f*x])/(3
*f*(a + a*Sec[e + f*x])^2)

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx &=-\frac {2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {\int \frac {\sec (e+f x) (-4 a c+3 a c \sec (e+f x))}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac {2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {c \int \sec (e+f x) \, dx}{a^2}+\frac {(7 c) \int \frac {\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=-\frac {c \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac {2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {7 c \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 0.46, size = 335, normalized size = 4.79 \[ \frac {c \sec \left (\frac {e}{2}\right ) \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (-6 \sin \left (e+\frac {f x}{2}\right )+10 \sin \left (e+\frac {3 f x}{2}\right )+3 \cos \left (e+\frac {3 f x}{2}\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 \cos \left (2 e+\frac {3 f x}{2}\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+9 \cos \left (\frac {f x}{2}\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 \cos \left (e+\frac {f x}{2}\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 \cos \left (e+\frac {3 f x}{2}\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-3 \cos \left (2 e+\frac {3 f x}{2}\right ) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+24 \sin \left (\frac {f x}{2}\right )\right )}{6 a^2 f (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Cos[(e + f*x)/2]*Sec[e/2]*Sec[e + f*x]^2*(3*Cos[e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 3
*Cos[2*e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 9*Cos[(f*x)/2]*(Log[Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 9*Cos[e + (f*x)/2]*(Log[Cos[(e + f*x)/2] - Sin[(e +
f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 3*Cos[e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]] - 3*Cos[2*e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 24*Sin[(f*x)/2] - 6*Sin[e + (f*x)/2
] + 10*Sin[e + (3*f*x)/2]))/(6*a^2*f*(1 + Sec[e + f*x])^2)

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fricas [A]  time = 0.47, size = 123, normalized size = 1.76 \[ -\frac {3 \, {\left (c \cos \left (f x + e\right )^{2} + 2 \, c \cos \left (f x + e\right ) + c\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (c \cos \left (f x + e\right )^{2} + 2 \, c \cos \left (f x + e\right ) + c\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (5 \, c \cos \left (f x + e\right ) + 7 \, c\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(c*cos(f*x + e)^2 + 2*c*cos(f*x + e) + c)*log(sin(f*x + e) + 1) - 3*(c*cos(f*x + e)^2 + 2*c*cos(f*x +
e) + c)*log(-sin(f*x + e) + 1) - 2*(5*c*cos(f*x + e) + 7*c)*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(
f*x + e) + a^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*(-c*1/2/a^2*ln(abs(tan((f*x+exp(1))/2)-1))+c*1/2/a^2*ln(a
bs(tan((f*x+exp(1))/2)+1))+(-4/3*tan((f*x+exp(1))/2)^3*c*a^4-8*tan((f*x+exp(1))/2)*c*a^4)*1/8/a^6)

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maple [A]  time = 0.71, size = 81, normalized size = 1.16 \[ \frac {c \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3 f \,a^{2}}+\frac {2 c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{f \,a^{2}}+\frac {c \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,a^{2}}-\frac {c \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/f*c/a^2*tan(1/2*e+1/2*f*x)^3+2/f*c/a^2*tan(1/2*e+1/2*f*x)+1/f*c/a^2*ln(tan(1/2*e+1/2*f*x)-1)-1/f*c/a^2*ln(
tan(1/2*e+1/2*f*x)+1)

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maxima [B]  time = 0.61, size = 144, normalized size = 2.06 \[ \frac {c {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(co
s(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + c*(3*sin(f*x + e)/(cos(f*x + e) +
1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

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mupad [B]  time = 1.69, size = 44, normalized size = 0.63 \[ \frac {c\,\left (6\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\right )}{3\,a^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(6*tan(e/2 + (f*x)/2) - 6*atanh(tan(e/2 + (f*x)/2)) + tan(e/2 + (f*x)/2)^3))/(3*a^2*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {c \left (\int \left (- \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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