3.220 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=89 \[ \frac{(c+5 d) (c-d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{d^2 \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac{(c-d)^2 \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.153959, antiderivative size = 149, normalized size of antiderivative = 1.67, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3987, 89, 78, 63, 217, 203} \[ \frac{(c+5 d) (c-d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{(c-d)^2 \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}+\frac{2 d^2 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - d)^2*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + (2*d^2*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec
[e + f*x])]]*Tan[e + f*x])/(a*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(c + 5*d)*Tan[e
+ f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

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

Mathematica [B]  time = 0.736878, size = 181, normalized size = 2.03 \[ -\frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \left (-4 \left (c^2+c d-2 d^2\right ) \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right ) \cos ^2\left (\frac{1}{2} (e+f x)\right )+(c-d)^2 \tan \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right )+(c-d)^2 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )+6 d^2 \cos ^3\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )}{3 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^2)/(a + a*Sec[e + f*x])^2,x]

[Out]

(-2*Cos[(e + f*x)/2]*(6*d^2*Cos[(e + f*x)/2]^3*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]]) + (c - d)^2*Sec[e/2]*Sin[(f*x)/2] - 4*(c^2 + c*d - 2*d^2)*Cos[(e + f*x)/2]^2*Sec[e/2]*S
in[(f*x)/2] + (c - d)^2*Cos[(e + f*x)/2]*Tan[e/2]))/(3*a^2*f*(1 + Cos[e + f*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.074, size = 170, normalized size = 1.9 \begin{align*}{\frac{{d}^{2}}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }+{\frac{cd}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{cd}{f{a}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }+{\frac{{c}^{2}}{2\,f{a}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{{d}^{2}}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-{\frac{{c}^{2}}{6\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{{d}^{2}}{6\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}-{\frac{3\,{d}^{2}}{2\,f{a}^{2}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^2,x)

[Out]

1/f/a^2*d^2*ln(tan(1/2*f*x+1/2*e)+1)+1/3/f/a^2*tan(1/2*f*x+1/2*e)^3*c*d+1/f/a^2*c*d*tan(1/2*f*x+1/2*e)+1/2/f/a
^2*tan(1/2*f*x+1/2*e)*c^2-1/f/a^2*d^2*ln(tan(1/2*f*x+1/2*e)-1)-1/6/f/a^2*tan(1/2*f*x+1/2*e)^3*c^2-1/6/f/a^2*ta
n(1/2*f*x+1/2*e)^3*d^2-3/2/f/a^2*tan(1/2*f*x+1/2*e)*d^2

________________________________________________________________________________________

Maxima [B]  time = 1.00558, size = 263, normalized size = 2.96 \begin{align*} -\frac{d^{2}{\left (\frac{\frac{9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac{2 \, c d{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac{c^{2}{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/6*(d^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) - 2*c*d*(3*sin(f*x + e)/(cos(f*x
+ e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^
3/(cos(f*x + e) + 1)^3)/a^2)/f

________________________________________________________________________________________

Fricas [A]  time = 0.481805, size = 383, normalized size = 4.3 \begin{align*} \frac{3 \,{\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (d^{2} \cos \left (f x + e\right )^{2} + 2 \, d^{2} \cos \left (f x + e\right ) + d^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (c^{2} + 4 \, c d - 5 \, d^{2} + 2 \,{\left (c^{2} + c d - 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/6*(3*(d^2*cos(f*x + e)^2 + 2*d^2*cos(f*x + e) + d^2)*log(sin(f*x + e) + 1) - 3*(d^2*cos(f*x + e)^2 + 2*d^2*c
os(f*x + e) + d^2)*log(-sin(f*x + e) + 1) + 2*(c^2 + 4*c*d - 5*d^2 + 2*(c^2 + c*d - 2*d^2)*cos(f*x + e))*sin(f
*x + e))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{2} \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} \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{2 c d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**2/(a+a*sec(f*x+e))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.35398, size = 224, normalized size = 2.52 \begin{align*} \frac{\frac{6 \, d^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, d^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{4} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a^{4} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a^{4} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, a^{4} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 \, a^{4} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

1/6*(6*d^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 6*d^2*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - (a^4*c^2*ta
n(1/2*f*x + 1/2*e)^3 - 2*a^4*c*d*tan(1/2*f*x + 1/2*e)^3 + a^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 3*a^4*c^2*tan(1/2*f
*x + 1/2*e) - 6*a^4*c*d*tan(1/2*f*x + 1/2*e) + 9*a^4*d^2*tan(1/2*f*x + 1/2*e))/a^6)/f