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

Optimal. Leaf size=78 \[ -\frac{a^3 \tan (e+f x)}{c f}+\frac{8 a^3 \cot (e+f x)}{c f}+\frac{8 a^3 \csc (e+f x)}{c f}-\frac{4 a^3 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{a^3 x}{c} \]

[Out]

(a^3*x)/c - (4*a^3*ArcTanh[Sin[e + f*x]])/(c*f) + (8*a^3*Cot[e + f*x])/(c*f) + (8*a^3*Csc[e + f*x])/(c*f) - (a
^3*Tan[e + f*x])/(c*f)

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Rubi [A]  time = 0.208792, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {3904, 3886, 3473, 8, 2606, 3767, 2621, 321, 207, 2620, 14} \[ -\frac{a^3 \tan (e+f x)}{c f}+\frac{8 a^3 \cot (e+f x)}{c f}+\frac{8 a^3 \csc (e+f x)}{c f}-\frac{4 a^3 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{a^3 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x)/c - (4*a^3*ArcTanh[Sin[e + f*x]])/(c*f) + (8*a^3*Cot[e + f*x])/(c*f) + (8*a^3*Csc[e + f*x])/(c*f) - (a
^3*Tan[e + f*x])/(c*f)

Rule 3904

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

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx &=-\frac{\int \cot ^2(e+f x) (a+a \sec (e+f x))^4 \, dx}{a c}\\ &=-\frac{\int \left (a^4 \cot ^2(e+f x)+4 a^4 \cot (e+f x) \csc (e+f x)+6 a^4 \csc ^2(e+f x)+4 a^4 \csc ^2(e+f x) \sec (e+f x)+a^4 \csc ^2(e+f x) \sec ^2(e+f x)\right ) \, dx}{a c}\\ &=-\frac{a^3 \int \cot ^2(e+f x) \, dx}{c}-\frac{a^3 \int \csc ^2(e+f x) \sec ^2(e+f x) \, dx}{c}-\frac{\left (4 a^3\right ) \int \cot (e+f x) \csc (e+f x) \, dx}{c}-\frac{\left (4 a^3\right ) \int \csc ^2(e+f x) \sec (e+f x) \, dx}{c}-\frac{\left (6 a^3\right ) \int \csc ^2(e+f x) \, dx}{c}\\ &=\frac{a^3 \cot (e+f x)}{c f}+\frac{a^3 \int 1 \, dx}{c}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1+x^2}{x^2} \, dx,x,\tan (e+f x)\right )}{c f}+\frac{\left (4 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\csc (e+f x))}{c f}+\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{c f}+\frac{\left (6 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (e+f x))}{c f}\\ &=\frac{a^3 x}{c}+\frac{7 a^3 \cot (e+f x)}{c f}+\frac{8 a^3 \csc (e+f x)}{c f}-\frac{a^3 \operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}\right ) \, dx,x,\tan (e+f x)\right )}{c f}+\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (e+f x)\right )}{c f}\\ &=\frac{a^3 x}{c}-\frac{4 a^3 \tanh ^{-1}(\sin (e+f x))}{c f}+\frac{8 a^3 \cot (e+f x)}{c f}+\frac{8 a^3 \csc (e+f x)}{c f}-\frac{a^3 \tan (e+f x)}{c f}\\ \end{align*}

Mathematica [B]  time = 2.3685, size = 240, normalized size = 3.08 \[ \frac{a^3 \cos ^2(e+f x) \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 \left (8 \csc \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \sec \left (\frac{1}{2} (e+f x)\right )+\tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}-4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-f x\right )\right )}{4 f (c-c \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[e + f*x]^2*Sec[(e + f*x)/2]^4*(1 + Sec[e + f*x])^3*Tan[(e + f*x)/2]*(8*Csc[e/2]*Sec[(e + f*x)/2]*Sin[
(f*x)/2] + (-(f*x) - 4*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 4*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] +
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]
+ Sin[(e + f*x)/2])))*Tan[(e + f*x)/2]))/(4*f*(c - c*Sec[e + f*x]))

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Maple [A]  time = 0.083, size = 137, normalized size = 1.8 \begin{align*} 2\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{fc}}+{\frac{{a}^{3}}{fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-4\,{\frac{{a}^{3}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }{fc}}+{\frac{{a}^{3}}{fc} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+4\,{\frac{{a}^{3}\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }{fc}}+8\,{\frac{{a}^{3}}{fc\tan \left ( 1/2\,fx+e/2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.55327, size = 370, normalized size = 4.74 \begin{align*} -\frac{a^{3}{\left (\frac{\frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} - a^{3}{\left (\frac{2 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} + \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} + 3 \, a^{3}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac{\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac{3 \, a^{3}{\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.09735, size = 313, normalized size = 4.01 \begin{align*} \frac{a^{3} f x \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 9 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) - a^{3}}{c f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*f*x*cos(f*x + e)*sin(f*x + e) - 2*a^3*cos(f*x + e)*log(sin(f*x + e) + 1)*sin(f*x + e) + 2*a^3*cos(f*x + e
)*log(-sin(f*x + e) + 1)*sin(f*x + e) + 9*a^3*cos(f*x + e)^2 + 8*a^3*cos(f*x + e) - a^3)/(c*f*cos(f*x + e)*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{3 \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{\sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} - 1}\, dx + \int \frac{1}{\sec{\left (e + f x \right )} - 1}\, dx\right )}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34976, size = 158, normalized size = 2.03 \begin{align*} \frac{\frac{{\left (f x + e\right )} a^{3}}{c} - \frac{4 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c} + \frac{4 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c} + \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} c}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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