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

Optimal. Leaf size=100 \[ \frac{10 c^3 \tan (e+f x)}{a f}-\frac{15 c^3 \tanh ^{-1}(\sin (e+f x))}{2 a f}-\frac{5 c^3 \tan (e+f x) \sec (e+f x)}{2 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^2}{f (a \sec (e+f x)+a)} \]

[Out]

(-15*c^3*ArcTanh[Sin[e + f*x]])/(2*a*f) + (10*c^3*Tan[e + f*x])/(a*f) - (5*c^3*Sec[e + f*x]*Tan[e + f*x])/(2*a
*f) + (2*c*(c - c*Sec[e + f*x])^2*Tan[e + f*x])/(f*(a + a*Sec[e + f*x]))

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Rubi [A]  time = 0.130097, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3957, 3788, 3767, 8, 4046, 3770} \[ \frac{10 c^3 \tan (e+f x)}{a f}-\frac{15 c^3 \tanh ^{-1}(\sin (e+f x))}{2 a f}-\frac{5 c^3 \tan (e+f x) \sec (e+f x)}{2 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^2}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*c^3*ArcTanh[Sin[e + f*x]])/(2*a*f) + (10*c^3*Tan[e + f*x])/(a*f) - (5*c^3*Sec[e + f*x]*Tan[e + f*x])/(2*a
*f) + (2*c*(c - c*Sec[e + f*x])^2*Tan[e + f*x])/(f*(a + a*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 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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)} \, dx &=\frac{2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(5 c) \int \sec (e+f x) (c-c \sec (e+f x))^2 \, dx}{a}\\ &=\frac{2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(5 c) \int \sec (e+f x) \left (c^2+c^2 \sec ^2(e+f x)\right ) \, dx}{a}+\frac{\left (10 c^3\right ) \int \sec ^2(e+f x) \, dx}{a}\\ &=-\frac{5 c^3 \sec (e+f x) \tan (e+f x)}{2 a f}+\frac{2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (15 c^3\right ) \int \sec (e+f x) \, dx}{2 a}-\frac{\left (10 c^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{a f}\\ &=-\frac{15 c^3 \tanh ^{-1}(\sin (e+f x))}{2 a f}+\frac{10 c^3 \tan (e+f x)}{a f}-\frac{5 c^3 \sec (e+f x) \tan (e+f x)}{2 a f}+\frac{2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end{align*}

Mathematica [B]  time = 3.17075, size = 287, normalized size = 2.87 \[ \frac{\cos ^2(e+f x) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right ) (c-c \sec (e+f x))^3 \left (\cot \left (\frac{1}{2} (e+f x)\right ) \left (-\frac{16 \sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{1}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-30 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+30 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-32 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \csc \left (\frac{1}{2} (e+f x)\right )\right )}{16 a f (\sec (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^2*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^4*(c - c*Sec[e + f*x])^3*(-32*Csc[(e + f*x)/2]*Sec[e/2]*Sin[
(f*x)/2] + Cot[(e + f*x)/2]*(-30*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 30*Log[Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]] + (Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2) - (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^(-2) - (16*Sin[
f*x])/((Cos[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])))))/(16*a*f*(1 + Sec[e + f*x]))

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Maple [A]  time = 0.078, size = 164, normalized size = 1.6 \begin{align*} 8\,{\frac{{c}^{3}\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}+{\frac{{c}^{3}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-2}}-{\frac{9\,{c}^{3}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{15\,{c}^{3}}{2\,fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{c}^{3}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-2}}-{\frac{9\,{c}^{3}}{2\,fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{15\,{c}^{3}}{2\,fa}\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)),x)

[Out]

8/f*c^3/a*tan(1/2*f*x+1/2*e)+1/2/f*c^3/a/(tan(1/2*f*x+1/2*e)+1)^2-9/2/f*c^3/a/(tan(1/2*f*x+1/2*e)+1)-15/2/f*c^
3/a*ln(tan(1/2*f*x+1/2*e)+1)-1/2/f*c^3/a/(tan(1/2*f*x+1/2*e)-1)^2-9/2/f*c^3/a/(tan(1/2*f*x+1/2*e)-1)+15/2/f*c^
3/a*ln(tan(1/2*f*x+1/2*e)-1)

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Maxima [B]  time = 1.01553, size = 521, normalized size = 5.21 \begin{align*} \frac{c^{3}{\left (\frac{2 \,{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a - \frac{2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} - \frac{3 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} + \frac{3 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} + \frac{2 \, \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 6 \, c^{3}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (f x + e\right )}{{\left (a - \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 6 \, c^{3}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + \frac{2 \, c^{3} \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.482304, size = 344, normalized size = 3.44 \begin{align*} -\frac{15 \,{\left (c^{3} \cos \left (f x + e\right )^{3} + c^{3} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (c^{3} \cos \left (f x + e\right )^{3} + c^{3} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (24 \, c^{3} \cos \left (f x + e\right )^{2} + 7 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \sin \left (f x + e\right )}{4 \,{\left (a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )^{2}\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)),x, algorithm="fricas")

[Out]

-1/4*(15*(c^3*cos(f*x + e)^3 + c^3*cos(f*x + e)^2)*log(sin(f*x + e) + 1) - 15*(c^3*cos(f*x + e)^3 + c^3*cos(f*
x + e)^2)*log(-sin(f*x + e) + 1) - 2*(24*c^3*cos(f*x + e)^2 + 7*c^3*cos(f*x + e) - c^3)*sin(f*x + e))/(a*f*cos
(f*x + e)^3 + a*f*cos(f*x + e)^2)

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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{\left (e + f x \right )} + 1}\, dx + \int \frac{3 \sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int - \frac{3 \sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx\right )}{a} \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)),x)

[Out]

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

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Giac [A]  time = 1.39285, size = 165, normalized size = 1.65 \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} - \frac{15 \, c^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{16 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{2 \,{\left (9 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 7 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{2} a}}{2 \, 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)),x, algorithm="giac")

[Out]

-1/2*(15*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 15*c^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a - 16*c^3*tan(1
/2*f*x + 1/2*e)/a + 2*(9*c^3*tan(1/2*f*x + 1/2*e)^3 - 7*c^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1
)^2*a))/f