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

Optimal. Leaf size=132 \[ -\frac{a^3 \tanh ^{-1}(\sin (e+f x))}{c^3 f}-\frac{2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )}+\frac{2 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 c f (c-c \sec (e+f x))^2}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.21442, antiderivative size = 132, 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} \[ -\frac{a^3 \tanh ^{-1}(\sin (e+f x))}{c^3 f}-\frac{2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )}+\frac{2 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 c f (c-c \sec (e+f x))^2}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.108118, size = 139, normalized size = 1.05 \[ -\frac{a^3 \left (-\frac{26 \cot \left (\frac{1}{2} (e+f x)\right )}{15 f}-\frac{2 \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right )}{5 f}+\frac{2 \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{15 f}-\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}+\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*((-26*Cot[(e + f*x)/2])/(15*f) + (2*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2)/(15*f) - (2*Cot[(e + f*x)/2]*C
sc[(e + f*x)/2]^4)/(5*f) - Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2
]]/f))/c^3)

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Maple [A]  time = 0.09, size = 113, normalized size = 0.9 \begin{align*} -{\frac{{a}^{3}}{f{c}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }+{\frac{{a}^{3}}{f{c}^{3}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }+{\frac{2\,{a}^{3}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-5}}+{\frac{2\,{a}^{3}}{3\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{3}}{f{c}^{3}\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.02943, size = 417, normalized size = 3.16 \begin{align*} -\frac{a^{3}{\left (\frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{3}} - \frac{60 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{3}} - \frac{{\left (\frac{20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} - \frac{3 \, a^{3}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac{a^{3}{\left (\frac{10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac{9 \, a^{3}{\left (\frac{5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(a^3*(60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^3
- (20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/
(c^3*sin(f*x + e)^5)) - 3*a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + 9*a^3*(5*sin(f*x + e)^4/(cos(f*x + e
) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5))/f

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Fricas [A]  time = 0.484952, size = 446, normalized size = 3.38 \begin{align*} \frac{52 \, a^{3} \cos \left (f x + e\right )^{3} - 44 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) + 92 \, a^{3} - 15 \,{\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 15 \,{\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right )}{30 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(52*a^3*cos(f*x + e)^3 - 44*a^3*cos(f*x + e)^2 - 4*a^3*cos(f*x + e) + 92*a^3 - 15*(a^3*cos(f*x + e)^2 - 2
*a^3*cos(f*x + e) + a^3)*log(sin(f*x + e) + 1)*sin(f*x + e) + 15*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^
3)*log(-sin(f*x + e) + 1)*sin(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{a^{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 )}{c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**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))/c**3

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Giac [A]  time = 1.28838, size = 144, normalized size = 1.09 \begin{align*} -\frac{\frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} - \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 5 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 15*a^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^3 - 2*(15*a
^3*tan(1/2*f*x + 1/2*e)^4 + 5*a^3*tan(1/2*f*x + 1/2*e)^2 + 3*a^3)/(c^3*tan(1/2*f*x + 1/2*e)^5))/f