3.246 \(\int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=115 \[ \frac{2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \tan (e+f x)}{3 f}+\frac{\left (a \left (2 c^2+d^2\right )+2 b c d\right ) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{d (3 a d+2 b c) \tan (e+f x) \sec (e+f x)}{6 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^2}{3 f} \]

[Out]

((2*b*c*d + a*(2*c^2 + d^2))*ArcTanh[Sin[e + f*x]])/(2*f) + (2*(3*a*c*d + b*(c^2 + d^2))*Tan[e + f*x])/(3*f) +
(d*(2*b*c + 3*a*d)*Sec[e + f*x]*Tan[e + f*x])/(6*f) + (b*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(3*f)

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Rubi [A]  time = 0.185081, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4002, 3997, 3787, 3770, 3767, 8} \[ \frac{2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \tan (e+f x)}{3 f}+\frac{\left (a \left (2 c^2+d^2\right )+2 b c d\right ) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{d (3 a d+2 b c) \tan (e+f x) \sec (e+f x)}{6 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c*d + a*(2*c^2 + d^2))*ArcTanh[Sin[e + f*x]])/(2*f) + (2*(3*a*c*d + b*(c^2 + d^2))*Tan[e + f*x])/(3*f) +
(d*(2*b*c + 3*a*d)*Sec[e + f*x]*Tan[e + f*x])/(6*f) + (b*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(3*f)

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.593982, size = 88, normalized size = 0.77 \[ \frac{3 \left (a \left (2 c^2+d^2\right )+2 b c d\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (3 d (a d+2 b c) \sec (e+f x)+12 a c d+6 b \left (c^2+d^2\right )+2 b d^2 \tan ^2(e+f x)\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(2*b*c*d + a*(2*c^2 + d^2))*ArcTanh[Sin[e + f*x]] + Tan[e + f*x]*(12*a*c*d + 6*b*(c^2 + d^2) + 3*d*(2*b*c +
a*d)*Sec[e + f*x] + 2*b*d^2*Tan[e + f*x]^2))/(6*f)

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Maple [A]  time = 0.035, size = 174, normalized size = 1.5 \begin{align*}{\frac{{c}^{2}a\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+2\,{\frac{acd\tan \left ( fx+e \right ) }{f}}+{\frac{a{d}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{a{d}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{{c}^{2}b\tan \left ( fx+e \right ) }{f}}+{\frac{bcd\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+{\frac{bcd\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{2\,{d}^{2}b\tan \left ( fx+e \right ) }{3\,f}}+{\frac{{d}^{2}b\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*c^2*a*ln(sec(f*x+e)+tan(f*x+e))+2/f*a*c*d*tan(f*x+e)+1/2/f*a*d^2*sec(f*x+e)*tan(f*x+e)+1/2/f*a*d^2*ln(sec(
f*x+e)+tan(f*x+e))+1/f*c^2*b*tan(f*x+e)+1/f*b*c*d*sec(f*x+e)*tan(f*x+e)+1/f*b*c*d*ln(sec(f*x+e)+tan(f*x+e))+2/
3/f*d^2*b*tan(f*x+e)+1/3/f*d^2*b*tan(f*x+e)*sec(f*x+e)^2

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Maxima [A]  time = 1.02573, size = 223, normalized size = 1.94 \begin{align*} \frac{4 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} b d^{2} - 6 \, b c d{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 3 \, a d^{2}{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 12 \, a c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 12 \, b c^{2} \tan \left (f x + e\right ) + 24 \, a c d \tan \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(f*x + e)^3 + 3*tan(f*x + e))*b*d^2 - 6*b*c*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x +
e) + 1) + log(sin(f*x + e) - 1)) - 3*a*d^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log
(sin(f*x + e) - 1)) + 12*a*c^2*log(sec(f*x + e) + tan(f*x + e)) + 12*b*c^2*tan(f*x + e) + 24*a*c*d*tan(f*x + e
))/f

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Fricas [A]  time = 0.560209, size = 371, normalized size = 3.23 \begin{align*} \frac{3 \,{\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, b d^{2} + 2 \,{\left (3 \, b c^{2} + 6 \, a c d + 2 \, b d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \, f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*a*c^2 + 2*b*c*d + a*d^2)*cos(f*x + e)^3*log(sin(f*x + e) + 1) - 3*(2*a*c^2 + 2*b*c*d + a*d^2)*cos(f
*x + e)^3*log(-sin(f*x + e) + 1) + 2*(2*b*d^2 + 2*(3*b*c^2 + 6*a*c*d + 2*b*d^2)*cos(f*x + e)^2 + 3*(2*b*c*d +
a*d^2)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(e + f*x))*(c + d*sec(e + f*x))**2*sec(e + f*x), x)

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Giac [B]  time = 1.34212, size = 419, normalized size = 3.64 \begin{align*} \frac{3 \,{\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, b c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 12 \, a c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 6 \, b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 \, a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 6 \, b d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 12 \, b c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 24 \, a c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 4 \, b d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, b c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, a c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, b d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(2*a*c^2 + 2*b*c*d + a*d^2)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*(2*a*c^2 + 2*b*c*d + a*d^2)*log(abs(
tan(1/2*f*x + 1/2*e) - 1)) - 2*(6*b*c^2*tan(1/2*f*x + 1/2*e)^5 + 12*a*c*d*tan(1/2*f*x + 1/2*e)^5 - 6*b*c*d*tan
(1/2*f*x + 1/2*e)^5 - 3*a*d^2*tan(1/2*f*x + 1/2*e)^5 + 6*b*d^2*tan(1/2*f*x + 1/2*e)^5 - 12*b*c^2*tan(1/2*f*x +
1/2*e)^3 - 24*a*c*d*tan(1/2*f*x + 1/2*e)^3 - 4*b*d^2*tan(1/2*f*x + 1/2*e)^3 + 6*b*c^2*tan(1/2*f*x + 1/2*e) +
12*a*c*d*tan(1/2*f*x + 1/2*e) + 6*b*c*d*tan(1/2*f*x + 1/2*e) + 3*a*d^2*tan(1/2*f*x + 1/2*e) + 6*b*d^2*tan(1/2*
f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^3)/f