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

Optimal. Leaf size=88 \[ \frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac {4 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}-\frac {8 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac {a^3 x}{c^2} \]

[Out]

a^3*x/c^2+a^3*arctanh(sin(f*x+e))/c^2/f-8/3*a^3*tan(f*x+e)/c^2/f/(1-sec(f*x+e))^2+4/3*a^3*tan(f*x+e)/c^2/f/(1-
sec(f*x+e))

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Rubi [A]  time = 0.36, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3903, 3777, 3919, 3794, 3796, 3797, 3799, 3998, 3770} \[ \frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}+\frac {4 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}-\frac {8 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac {a^3 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3770

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

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3796

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3797

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(2*m + 1)), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3903

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dis
t[c^n, Int[ExpandTrig[(1 + (d*csc[e + f*x])/c)^n, (a + b*csc[e + f*x])^m, x], x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx &=\frac {\int \left (\frac {a^3}{(1-\sec (e+f x))^2}+\frac {3 a^3 \sec (e+f x)}{(1-\sec (e+f x))^2}+\frac {3 a^3 \sec ^2(e+f x)}{(1-\sec (e+f x))^2}+\frac {a^3 \sec ^3(e+f x)}{(1-\sec (e+f x))^2}\right ) \, dx}{c^2}\\ &=\frac {a^3 \int \frac {1}{(1-\sec (e+f x))^2} \, dx}{c^2}+\frac {a^3 \int \frac {\sec ^3(e+f x)}{(1-\sec (e+f x))^2} \, dx}{c^2}+\frac {\left (3 a^3\right ) \int \frac {\sec (e+f x)}{(1-\sec (e+f x))^2} \, dx}{c^2}+\frac {\left (3 a^3\right ) \int \frac {\sec ^2(e+f x)}{(1-\sec (e+f x))^2} \, dx}{c^2}\\ &=-\frac {8 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}-\frac {a^3 \int \frac {-3-\sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}+\frac {a^3 \int \frac {(-2-3 \sec (e+f x)) \sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}+\frac {a^3 \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{c^2}-\frac {\left (2 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{c^2}\\ &=\frac {a^3 x}{c^2}-\frac {8 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac {a^3 \tan (e+f x)}{c^2 f (1-\sec (e+f x))}+\frac {a^3 \int \sec (e+f x) \, dx}{c^2}+\frac {\left (4 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}-\frac {\left (5 a^3\right ) \int \frac {\sec (e+f x)}{1-\sec (e+f x)} \, dx}{3 c^2}\\ &=\frac {a^3 x}{c^2}+\frac {a^3 \tanh ^{-1}(\sin (e+f x))}{c^2 f}-\frac {8 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))^2}+\frac {4 a^3 \tan (e+f x)}{3 c^2 f (1-\sec (e+f x))}\\ \end {align*}

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Mathematica [B]  time = 1.18, size = 177, normalized size = 2.01 \[ \frac {a^3 (\cos (e+f x)+1)^3 \tan \left (\frac {1}{2} (e+f x)\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \left (-4 \cot \left (\frac {e}{2}\right ) \tan \left (\frac {1}{2} (e+f x)\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )+4 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sec \left (\frac {1}{2} (e+f x)\right )+3 \tan ^3\left (\frac {1}{2} (e+f x)\right ) \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+f x\right )\right )}{6 c^2 f (\cos (e+f x)-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(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]*(4*Csc[e/2]*Sec[(e + f*x)/2]*Sin[(f*x)/2] - 4*Co
t[e/2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] + 3*(f*x - Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Log[Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]])*Tan[(e + f*x)/2]^3))/(6*c^2*f*(-1 + Cos[e + f*x])^2)

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fricas [A]  time = 0.45, size = 156, normalized size = 1.77 \[ \frac {8 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} \cos \left (f x + e\right ) + 8 \, a^{3} + 3 \, {\left (a^{3} \cos \left (f x + e\right ) - a^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{3} \cos \left (f x + e\right ) - a^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 6 \, {\left (a^{3} f x \cos \left (f x + e\right ) - a^{3} f x\right )} \sin \left (f x + e\right )}{6 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )} \]

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))^2,x, algorithm="fricas")

[Out]

1/6*(8*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 8*a^3 + 3*(a^3*cos(f*x + e) - a^3)*log(sin(f*x + e) + 1)*sin
(f*x + e) - 3*(a^3*cos(f*x + e) - a^3)*log(-sin(f*x + e) + 1)*sin(f*x + e) + 6*(a^3*f*x*cos(f*x + e) - a^3*f*x
)*sin(f*x + e))/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e))

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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))^2,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*(-a^3*1/2/c^2*ln(abs(tan((f*x+exp(1))/2)-1))+a^3*1/2/c^2*l
n(abs(tan((f*x+exp(1))/2)+1))+2*a^3*1/2/c^2*(f*x+exp(1))/2-2*a^3*1/3/c^2/tan((f*x+exp(1))/2)^3)

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maple [A]  time = 0.83, size = 90, normalized size = 1.02 \[ -\frac {4 a^{3}}{3 f \,c^{2} \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}-\frac {a^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,c^{2}}+\frac {a^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,c^{2}}+\frac {2 a^{3} \arctan \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f \,c^{2}} \]

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))^2,x)

[Out]

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

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maxima [B]  time = 0.44, size = 274, normalized size = 3.11 \[ \frac {a^{3} {\left (\frac {12 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{2}} + \frac {{\left (\frac {9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} + a^{3} {\left (\frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}} - \frac {{\left (\frac {9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} - \frac {3 \, a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}} + \frac {3 \, a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}}{6 \, f} \]

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))^2,x, algorithm="maxima")

[Out]

1/6*(a^3*(12*arctan(sin(f*x + e)/(cos(f*x + e) + 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)) + 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) + 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

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mupad [B]  time = 1.44, size = 45, normalized size = 0.51 \[ \frac {a^3\,x}{c^2}+\frac {a^3\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {4\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}\right )}{c^2\,f} \]

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))^2,x)

[Out]

(a^3*x)/c^2 + (a^3*(2*atanh(tan(e/2 + (f*x)/2)) - (4*cot(e/2 + (f*x)/2)^3)/3))/(c^2*f)

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

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))**2,x)

[Out]

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

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