3.28 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=119 \[ \frac{5 a^3 \tan (e+f x)}{c^2 f}+\frac{5 a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac{10 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 f \left (c^2-c^2 \sec (e+f x)\right )}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2} \]

[Out]

(5*a^3*ArcTanh[Sin[e + f*x]])/(c^2*f) + (5*a^3*Tan[e + f*x])/(c^2*f) - (2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x
])/(3*f*(c - c*Sec[e + f*x])^2) + (10*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*(c^2 - c^2*Sec[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.183051, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {3957, 3787, 3770, 3767, 8} \[ \frac{5 a^3 \tan (e+f x)}{c^2 f}+\frac{5 a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac{10 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{3 f \left (c^2-c^2 \sec (e+f x)\right )}-\frac{2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^2,x]

[Out]

(5*a^3*ArcTanh[Sin[e + f*x]])/(c^2*f) + (5*a^3*Tan[e + f*x])/(c^2*f) - (2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x
])/(3*f*(c - c*Sec[e + f*x])^2) + (10*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*(c^2 - c^2*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 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 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx &=-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}-\frac{(5 a) \int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{c-c \sec (e+f x)} \, dx}{3 c}\\ &=-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}+\frac{10 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (c^2-c^2 \sec (e+f x)\right )}+\frac{\left (5 a^2\right ) \int \sec (e+f x) (a+a \sec (e+f x)) \, dx}{c^2}\\ &=-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}+\frac{10 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (c^2-c^2 \sec (e+f x)\right )}+\frac{\left (5 a^3\right ) \int \sec (e+f x) \, dx}{c^2}+\frac{\left (5 a^3\right ) \int \sec ^2(e+f x) \, dx}{c^2}\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}+\frac{10 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (c^2-c^2 \sec (e+f x)\right )}-\frac{\left (5 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{c^2 f}\\ &=\frac{5 a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac{5 a^3 \tan (e+f x)}{c^2 f}-\frac{2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}+\frac{10 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (c^2-c^2 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 3.24675, size = 402, normalized size = 3.38 \[ \frac{a^3 (\cos (e+f x)+1)^3 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (-48 \sin (e) \csc ^3\left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sin ^7\left (\frac{1}{2} (e+f x)\right ) \csc ^4(e+f x)+\frac{1}{16} \csc ^3\left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (120 \cos (e+f x)-46 \cos (2 (e+f x))-76 \cos (2 e+f x)+23 \cos (e+2 f x)+23 \cos (3 e+2 f x)+42 \cos (e)-76 \cos (f x)-74) \sec ^5\left (\frac{1}{2} (e+f x)\right )-4 \cot ^2\left (\frac{e}{2}\right ) \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) (5 \cos (e+f x)-1) \tan ^2\left (\frac{1}{2} (e+f x)\right ) \sec ^3\left (\frac{1}{2} (e+f x)\right )+\cos (e) \csc ^2\left (\frac{e}{2}\right ) \cos (e+f x) \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^4\left (\frac{1}{2} (e+f x)\right ) \left (15 \sin ^2\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+4 \cot \left (\frac{e}{2}\right )\right )\right )}{6 c^2 f \left (\cot \left (\frac{e}{2}\right )-1\right ) \left (\cot \left (\frac{e}{2}\right )+1\right ) (\cos (e+f x)-1)^2 \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right ) \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^2,x]

[Out]

(a^3*(1 + Cos[e + f*x])^3*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*(((-74 + 42*Cos[e] - 76*Cos[f*x] + 120*Cos[e + f
*x] - 46*Cos[2*(e + f*x)] - 76*Cos[2*e + f*x] + 23*Cos[e + 2*f*x] + 23*Cos[3*e + 2*f*x])*Csc[e/2]^3*Sec[(e + f
*x)/2]^5*Sin[(f*x)/2])/16 - 48*Csc[e/2]^3*Csc[e + f*x]^4*Sin[e]*Sin[(f*x)/2]*Sin[(e + f*x)/2]^7 + Cos[e]*Cos[e
+ f*x]*Csc[e/2]^2*Sec[(e + f*x)/2]^4*(4*Cot[e/2] + 15*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2]])*Sin[(e + f*x)/2]^2)*Tan[(e + f*x)/2] - 4*(-1 + 5*Cos[e + f*x])*Cot[e/2]^2*Csc[e
/2]*Sec[(e + f*x)/2]^3*Sin[(f*x)/2]*Tan[(e + f*x)/2]^2))/(6*c^2*f*(-1 + Cos[e + f*x])^2*(-1 + Cot[e/2])*(1 + C
ot[e/2])*(-1 + Tan[(e + f*x)/2])*(1 + Tan[(e + f*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 0.078, size = 140, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}}{f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}+5\,{\frac{{a}^{3}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{f{c}^{2}}}-{\frac{{a}^{3}}{f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}-5\,{\frac{{a}^{3}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{f{c}^{2}}}-{\frac{4\,{a}^{3}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}-8\,{\frac{{a}^{3}}{f{c}^{2}\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))^2,x)

[Out]

-1/f*a^3/c^2/(tan(1/2*f*x+1/2*e)+1)+5/f*a^3/c^2*ln(tan(1/2*f*x+1/2*e)+1)-1/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)-5/
f*a^3/c^2*ln(tan(1/2*f*x+1/2*e)-1)-4/3/f*a^3/c^2/tan(1/2*f*x+1/2*e)^3-8/f*a^3/c^2/tan(1/2*f*x+1/2*e)

________________________________________________________________________________________

Maxima [B]  time = 1.03476, size = 471, normalized size = 3.96 \begin{align*} -\frac{a^{3}{\left (\frac{\frac{14 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{27 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 1}{\frac{c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac{12 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} + \frac{12 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}}\right )} - 3 \, a^{3}{\left (\frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} - \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}} - \frac{{\left (\frac{9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} + \frac{3 \, a^{3}{\left (\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}} - \frac{a^{3}{\left (\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

-1/6*(a^3*((14*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 27*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1)/(c^2*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 - c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 12*log(sin(f*x + e)/(cos(f*x + e) + 1
) + 1)/c^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^2) - 3*a^3*(6*log(sin(f*x + e)/(cos(f*x + e) + 1) +
1)/c^2 - 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^2 - (9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)*(cos(f*
x + e) + 1)^3/(c^2*sin(f*x + e)^3)) + 3*a^3*(3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)*(cos(f*x + e) + 1)^3/(
c^2*sin(f*x + e)^3) - a^3*(3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)*(cos(f*x + e) + 1)^3/(c^2*sin(f*x + e)^3
))/f

________________________________________________________________________________________

Fricas [A]  time = 0.48671, size = 410, normalized size = 3.45 \begin{align*} -\frac{46 \, a^{3} \cos \left (f x + e\right )^{3} - 22 \, a^{3} \cos \left (f x + e\right )^{2} - 62 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3} - 15 \,{\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3} \cos \left (f x + e\right )\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} - a^{3} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right )}{6 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right )\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))^2,x, algorithm="fricas")

[Out]

-1/6*(46*a^3*cos(f*x + e)^3 - 22*a^3*cos(f*x + e)^2 - 62*a^3*cos(f*x + e) + 6*a^3 - 15*(a^3*cos(f*x + e)^2 - a
^3*cos(f*x + e))*log(sin(f*x + e) + 1)*sin(f*x + e) + 15*(a^3*cos(f*x + e)^2 - a^3*cos(f*x + e))*log(-sin(f*x
+ e) + 1)*sin(f*x + e))/((c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e))*sin(f*x + e))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.32433, size = 165, normalized size = 1.39 \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^{2}} - \frac{15 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{6 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac{4 \,{\left (6 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a^{3}\right )}}{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{3 \, 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))^2,x, algorithm="giac")

[Out]

1/3*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^2 - 15*a^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/c^2 - 6*a^3*tan
(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2*e)^2 - 1)*c^2) - 4*(6*a^3*tan(1/2*f*x + 1/2*e)^2 + a^3)/(c^2*tan(1/2*f*x
+ 1/2*e)^3))/f