3.55 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=131 \[ -\frac{c^3 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{2 c^3 \tan (e+f x)}{f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{2 \tan (e+f x) \left (c^3-c^3 \sec (e+f x)\right )}{3 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

-((c^3*ArcTanh[Sin[e + f*x]])/(a^3*f)) + (2*c^3*Tan[e + f*x])/(f*(a^3 + a^3*Sec[e + f*x])) + (2*c*(c - c*Sec[e
+ f*x])^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) - (2*(c^3 - c^3*Sec[e + f*x])*Tan[e + f*x])/(3*a*f*(a +
a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.214818, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3957, 3770} \[ -\frac{c^3 \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac{2 c^3 \tan (e+f x)}{f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{2 \tan (e+f x) \left (c^3-c^3 \sec (e+f x)\right )}{3 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^2}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*ArcTanh[Sin[e + f*x]])/(a^3*f)) + (2*c^3*Tan[e + f*x])/(f*(a^3 + a^3*Sec[e + f*x])) + (2*c*(c - c*Sec[e
+ f*x])^2*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) - (2*(c^3 - c^3*Sec[e + f*x])*Tan[e + f*x])/(3*a*f*(a +
a*Sec[e + f*x])^2)

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3770

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

Rubi steps

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

Mathematica [A]  time = 0.119014, size = 139, normalized size = 1.06 \[ -\frac{c^3 \left (-\frac{26 \tan \left (\frac{1}{2} (e+f x)\right )}{15 f}-\frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^4\left (\frac{1}{2} (e+f x)\right )}{5 f}+\frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{15 f}-\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*(-(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f) + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]/f - (26*Tan[(
e + f*x)/2])/(15*f) + (2*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(15*f) - (2*Sec[(e + f*x)/2]^4*Tan[(e + f*x)/2])
/(5*f)))/a^3)

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Maple [A]  time = 0.084, size = 111, normalized size = 0.9 \begin{align*}{\frac{2\,{c}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}+{\frac{2\,{c}^{3}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+2\,{\frac{{c}^{3}\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{3}}}+{\frac{{c}^{3}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-{\frac{{c}^{3}}{f{a}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^3/(a+a*sec(f*x+e))^3,x)

[Out]

2/5/f*c^3/a^3*tan(1/2*f*x+1/2*e)^5+2/3/f*c^3/a^3*tan(1/2*f*x+1/2*e)^3+2/f*c^3/a^3*tan(1/2*f*x+1/2*e)+1/f*c^3/a
^3*ln(tan(1/2*f*x+1/2*e)-1)-1/f*c^3/a^3*ln(tan(1/2*f*x+1/2*e)+1)

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Maxima [B]  time = 1.02398, size = 410, normalized size = 3.13 \begin{align*} \frac{c^{3}{\left (\frac{\frac{105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac{3 \, c^{3}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac{c^{3}{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{9 \, c^{3}{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(c^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(c
os(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e)
+ 1) - 1)/a^3) + 3*c^3*(15*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*
x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + c^3*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 9*c^3*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)
^5/(cos(f*x + e) + 1)^5)/a^3)/f

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Fricas [A]  time = 0.485504, size = 479, normalized size = 3.66 \begin{align*} -\frac{15 \,{\left (c^{3} \cos \left (f x + e\right )^{3} + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 3 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (c^{3} \cos \left (f x + e\right )^{3} + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 3 \, c^{3} \cos \left (f x + e\right ) + c^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 4 \,{\left (13 \, c^{3} \cos \left (f x + e\right )^{2} + 24 \, c^{3} \cos \left (f x + e\right ) + 23 \, c^{3}\right )} \sin \left (f x + e\right )}{30 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(c^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*log(sin(f*x + e) + 1) - 15*(c
^3*cos(f*x + e)^3 + 3*c^3*cos(f*x + e)^2 + 3*c^3*cos(f*x + e) + c^3)*log(-sin(f*x + e) + 1) - 4*(13*c^3*cos(f*
x + e)^2 + 24*c^3*cos(f*x + e) + 23*c^3)*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*
f*cos(f*x + e) + a^3*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{c^{3} \left (\int - \frac{\sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**3*(Integral(-sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(3*sec(
e + f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(-3*sec(e + f*x)**3/(sec(
e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**4/(sec(e + f*x)**3 + 3*sec(
e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3

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Giac [A]  time = 1.26328, size = 154, normalized size = 1.18 \begin{align*} -\frac{\frac{15 \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac{15 \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac{2 \,{\left (3 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 5 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{12} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{a^{15}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 15*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 - 2*(3*a^
12*c^3*tan(1/2*f*x + 1/2*e)^5 + 5*a^12*c^3*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^3*tan(1/2*f*x + 1/2*e))/a^15)/f