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[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]))

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Rubi [A]  time = 0.361025, antiderivative size = 88, normalized size of antiderivative = 1., 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 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 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 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 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 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]

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{(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*}

Mathematica [B]  time = 1.17021, 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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Maple [A]  time = 0.087, size = 90, normalized size = 1. \begin{align*} -{\frac{4\,{a}^{3}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-3}}+2\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}}+{\frac{{a}^{3}}{f{c}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{a}^{3}}{f{c}^{2}}\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((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*f*x+1/2*e)^3+2/f*a^3/c^2*arctan(tan(1/2*f*x+1/2*e))+1/f*a^3/c^2*ln(tan(1/2*f*x+1/2*e)+1
)-1/f*a^3/c^2*ln(tan(1/2*f*x+1/2*e)-1)

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Maxima [B]  time = 1.53255, size = 370, normalized size = 4.2 \begin{align*} \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} \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))^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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Fricas [A]  time = 1.12544, size = 379, normalized size = 4.31 \begin{align*} \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 )} \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))^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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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 ^{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}} \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))**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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Giac [A]  time = 1.34594, size = 113, normalized size = 1.28 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} a^{3}}{c^{2}} + \frac{3 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac{3 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac{4 \, a^{3}}{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((a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="giac")

[Out]

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