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(f*x+e))/c/f+8*a^3*cot(f*x+e)/c/f+8*a^3*csc(f*x+e)/c/f-a^3*tan(f*x+e)/c/f

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 8

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

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]]

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 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 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 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 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 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 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 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 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])

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 = 2.58, 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]))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 125, normalized size = 1.60 \[ \frac {a^{3} f x \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 9 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) - a^{3}}{c f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]

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))

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*(-2*a^3/c*1/2*(f*x+exp(1))/2-2*a^3/c*ln(abs(tan((f*x+exp(
1))/2)-1))+2*a^3/c*ln(abs(tan((f*x+exp(1))/2)+1))+(-5*tan((f*x+exp(1))/2)^2*a^3+4*a^3)/c/(tan((f*x+exp(1))/2)^
3-tan((f*x+exp(1))/2)))

________________________________________________________________________________________

maple [A]  time = 0.70, size = 137, normalized size = 1.76 \[ \frac {8 a^{3}}{f c \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}+\frac {a^{3}}{f c \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}+\frac {4 a^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f c}+\frac {a^{3}}{f c \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}-\frac {4 a^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f c}+\frac {2 a^{3} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f c} \]

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]

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

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 274, normalized size = 3.51 \[ -\frac {a^{3} {\left (\frac {\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} - a^{3} {\left (\frac {2 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} + \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} + 3 \, a^{3} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac {3 \, a^{3} {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{f} \]

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

________________________________________________________________________________________

mupad [B]  time = 1.48, size = 85, normalized size = 1.09 \[ \frac {a^3\,x}{c}-\frac {10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-8\,a^3}{f\,\left (c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\right )}-\frac {8\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*x)/c - (10*a^3*tan(e/2 + (f*x)/2)^2 - 8*a^3)/(f*(c*tan(e/2 + (f*x)/2) - c*tan(e/2 + (f*x)/2)^3)) - (8*a^3
*atanh(tan(e/2 + (f*x)/2)))/(c*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {a^{3} \left (\int \frac {3 \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec {\left (e + f x \right )} - 1}\, dx\right )}{c} \]

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

________________________________________________________________________________________