3.245 \(\int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=180 \[ \frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \tan (e+f x)}{6 f}+\frac{\left (8 a c^3+12 a c d^2+12 b c^2 d+3 b d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{d \left (20 a c d+6 b c^2+9 b d^2\right ) \tan (e+f x) \sec (e+f x)}{24 f}+\frac{(4 a d+3 b c) \tan (e+f x) (c+d \sec (e+f x))^2}{12 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^3}{4 f} \]

[Out]

((8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*ArcTanh[Sin[e + f*x]])/(8*f) + ((4*a*d*(4*c^2 + d^2) + 3*b*(c^3
+ 4*c*d^2))*Tan[e + f*x])/(6*f) + (d*(6*b*c^2 + 20*a*c*d + 9*b*d^2)*Sec[e + f*x]*Tan[e + f*x])/(24*f) + ((3*b
*c + 4*a*d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(12*f) + (b*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(4*f)

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Rubi [A]  time = 0.356296, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \tan (e+f x)}{6 f}+\frac{\left (8 a c^3+12 a c d^2+12 b c^2 d+3 b d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{d \left (20 a c d+6 b c^2+9 b d^2\right ) \tan (e+f x) \sec (e+f x)}{24 f}+\frac{(4 a d+3 b c) \tan (e+f x) (c+d \sec (e+f x))^2}{12 f}+\frac{b \tan (e+f x) (c+d \sec (e+f x))^3}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + b*Sec[e + f*x])*(c + d*Sec[e + f*x])^3,x]

[Out]

((8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*ArcTanh[Sin[e + f*x]])/(8*f) + ((4*a*d*(4*c^2 + d^2) + 3*b*(c^3
+ 4*c*d^2))*Tan[e + f*x])/(6*f) + (d*(6*b*c^2 + 20*a*c*d + 9*b*d^2)*Sec[e + f*x]*Tan[e + f*x])/(24*f) + ((3*b
*c + 4*a*d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(12*f) + (b*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(4*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))^3 \, dx &=\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{4} \int \sec (e+f x) (c+d \sec (e+f x))^2 (4 a c+3 b d+(3 b c+4 a d) \sec (e+f x)) \, dx\\ &=\frac{(3 b c+4 a d) (c+d \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{12} \int \sec (e+f x) (c+d \sec (e+f x)) \left (12 a c^2+15 b c d+8 a d^2+\left (6 b c^2+20 a c d+9 b d^2\right ) \sec (e+f x)\right ) \, dx\\ &=\frac{d \left (6 b c^2+20 a c d+9 b d^2\right ) \sec (e+f x) \tan (e+f x)}{24 f}+\frac{(3 b c+4 a d) (c+d \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{24} \int \sec (e+f x) \left (3 \left (3 b d \left (4 c^2+d^2\right )+4 a \left (2 c^3+3 c d^2\right )\right )+4 \left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \sec (e+f x)\right ) \, dx\\ &=\frac{d \left (6 b c^2+20 a c d+9 b d^2\right ) \sec (e+f x) \tan (e+f x)}{24 f}+\frac{(3 b c+4 a d) (c+d \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}+\frac{1}{8} \left (8 a c^3+12 b c^2 d+12 a c d^2+3 b d^3\right ) \int \sec (e+f x) \, dx+\frac{1}{6} \left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \int \sec ^2(e+f x) \, dx\\ &=\frac{\left (8 a c^3+12 b c^2 d+12 a c d^2+3 b d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{d \left (6 b c^2+20 a c d+9 b d^2\right ) \sec (e+f x) \tan (e+f x)}{24 f}+\frac{(3 b c+4 a d) (c+d \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}-\frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{6 f}\\ &=\frac{\left (8 a c^3+12 b c^2 d+12 a c d^2+3 b d^3\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \tan (e+f x)}{6 f}+\frac{d \left (6 b c^2+20 a c d+9 b d^2\right ) \sec (e+f x) \tan (e+f x)}{24 f}+\frac{(3 b c+4 a d) (c+d \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{b (c+d \sec (e+f x))^3 \tan (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 1.08466, size = 143, normalized size = 0.79 \[ \frac{3 \left (4 a \left (2 c^3+3 c d^2\right )+3 b d \left (4 c^2+d^2\right )\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \left (8 \left (d^2 (a d+3 b c) \tan ^2(e+f x)+3 a d \left (3 c^2+d^2\right )+3 b \left (c^3+3 c d^2\right )\right )+9 d \left (4 a c d+b \left (4 c^2+d^2\right )\right ) \sec (e+f x)+6 b d^3 \sec ^3(e+f x)\right )}{24 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + b*Sec[e + f*x])*(c + d*Sec[e + f*x])^3,x]

[Out]

(3*(3*b*d*(4*c^2 + d^2) + 4*a*(2*c^3 + 3*c*d^2))*ArcTanh[Sin[e + f*x]] + Tan[e + f*x]*(9*d*(4*a*c*d + b*(4*c^2
+ d^2))*Sec[e + f*x] + 6*b*d^3*Sec[e + f*x]^3 + 8*(3*a*d*(3*c^2 + d^2) + 3*b*(c^3 + 3*c*d^2) + d^2*(3*b*c + a
*d)*Tan[e + f*x]^2)))/(24*f)

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Maple [A]  time = 0.04, size = 290, normalized size = 1.6 \begin{align*}{\frac{a{c}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+3\,{\frac{{c}^{2}da\tan \left ( fx+e \right ) }{f}}+{\frac{3\,ac{d}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{3\,ac{d}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{2\,{d}^{3}a\tan \left ( fx+e \right ) }{3\,f}}+{\frac{{d}^{3}a\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{{c}^{3}b\tan \left ( fx+e \right ) }{f}}+{\frac{3\,b{c}^{2}d\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{3\,b{c}^{2}d\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+2\,{\frac{{d}^{2}cb\tan \left ( fx+e \right ) }{f}}+{\frac{{d}^{2}cb\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{b{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{3\,b{d}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,b{d}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,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))^3,x)

[Out]

1/f*a*c^3*ln(sec(f*x+e)+tan(f*x+e))+3/f*a*c^2*d*tan(f*x+e)+3/2/f*a*d^2*c*sec(f*x+e)*tan(f*x+e)+3/2/f*a*d^2*c*l
n(sec(f*x+e)+tan(f*x+e))+2/3/f*a*d^3*tan(f*x+e)+1/3/f*a*d^3*tan(f*x+e)*sec(f*x+e)^2+1/f*c^3*b*tan(f*x+e)+3/2/f
*b*c^2*d*sec(f*x+e)*tan(f*x+e)+3/2/f*b*c^2*d*ln(sec(f*x+e)+tan(f*x+e))+2/f*d^2*c*b*tan(f*x+e)+1/f*d^2*c*b*tan(
f*x+e)*sec(f*x+e)^2+1/4/f*b*d^3*tan(f*x+e)*sec(f*x+e)^3+3/8/f*b*d^3*sec(f*x+e)*tan(f*x+e)+3/8/f*b*d^3*ln(sec(f
*x+e)+tan(f*x+e))

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Maxima [A]  time = 1.16779, size = 359, normalized size = 1.99 \begin{align*} \frac{48 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} b c d^{2} + 16 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a d^{3} - 3 \, b d^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 36 \, b c^{2} 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 )} - 36 \, a c 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 )} + 48 \, a c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 48 \, b c^{3} \tan \left (f x + e\right ) + 144 \, a c^{2} d \tan \left (f x + e\right )}{48 \, 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))^3,x, algorithm="maxima")

[Out]

1/48*(48*(tan(f*x + e)^3 + 3*tan(f*x + e))*b*c*d^2 + 16*(tan(f*x + e)^3 + 3*tan(f*x + e))*a*d^3 - 3*b*d^3*(2*(
3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(s
in(f*x + e) - 1)) - 36*b*c^2*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e)
- 1)) - 36*a*c*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)) + 48
*a*c^3*log(sec(f*x + e) + tan(f*x + e)) + 48*b*c^3*tan(f*x + e) + 144*a*c^2*d*tan(f*x + e))/f

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Fricas [A]  time = 0.578471, size = 510, normalized size = 2.83 \begin{align*} \frac{3 \,{\left (8 \, a c^{3} + 12 \, b c^{2} d + 12 \, a c d^{2} + 3 \, b d^{3}\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (8 \, a c^{3} + 12 \, b c^{2} d + 12 \, a c d^{2} + 3 \, b d^{3}\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (6 \, b d^{3} + 8 \,{\left (3 \, b c^{3} + 9 \, a c^{2} d + 6 \, b c d^{2} + 2 \, a d^{3}\right )} \cos \left (f x + e\right )^{3} + 9 \,{\left (4 \, b c^{2} d + 4 \, a c d^{2} + b d^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \,{\left (3 \, b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \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))^3,x, algorithm="fricas")

[Out]

1/48*(3*(8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 3*(8*a*c^3 + 12*b
*c^2*d + 12*a*c*d^2 + 3*b*d^3)*cos(f*x + e)^4*log(-sin(f*x + e) + 1) + 2*(6*b*d^3 + 8*(3*b*c^3 + 9*a*c^2*d + 6
*b*c*d^2 + 2*a*d^3)*cos(f*x + e)^3 + 9*(4*b*c^2*d + 4*a*c*d^2 + b*d^3)*cos(f*x + e)^2 + 8*(3*b*c*d^2 + a*d^3)*
cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4)

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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 )^{3} \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))**3,x)

[Out]

Integral((a + b*sec(e + f*x))*(c + d*sec(e + f*x))**3*sec(e + f*x), x)

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Giac [B]  time = 1.25672, size = 833, normalized size = 4.63 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="giac")

[Out]

1/24*(3*(8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*(8*a*c^3 + 12*b*c
^2*d + 12*a*c*d^2 + 3*b*d^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(24*b*c^3*tan(1/2*f*x + 1/2*e)^7 + 72*a*c^
2*d*tan(1/2*f*x + 1/2*e)^7 - 36*b*c^2*d*tan(1/2*f*x + 1/2*e)^7 - 36*a*c*d^2*tan(1/2*f*x + 1/2*e)^7 + 72*b*c*d^
2*tan(1/2*f*x + 1/2*e)^7 + 24*a*d^3*tan(1/2*f*x + 1/2*e)^7 - 15*b*d^3*tan(1/2*f*x + 1/2*e)^7 - 72*b*c^3*tan(1/
2*f*x + 1/2*e)^5 - 216*a*c^2*d*tan(1/2*f*x + 1/2*e)^5 + 36*b*c^2*d*tan(1/2*f*x + 1/2*e)^5 + 36*a*c*d^2*tan(1/2
*f*x + 1/2*e)^5 - 120*b*c*d^2*tan(1/2*f*x + 1/2*e)^5 - 40*a*d^3*tan(1/2*f*x + 1/2*e)^5 - 9*b*d^3*tan(1/2*f*x +
1/2*e)^5 + 72*b*c^3*tan(1/2*f*x + 1/2*e)^3 + 216*a*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 36*b*c^2*d*tan(1/2*f*x + 1/
2*e)^3 + 36*a*c*d^2*tan(1/2*f*x + 1/2*e)^3 + 120*b*c*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*a*d^3*tan(1/2*f*x + 1/2*e
)^3 - 9*b*d^3*tan(1/2*f*x + 1/2*e)^3 - 24*b*c^3*tan(1/2*f*x + 1/2*e) - 72*a*c^2*d*tan(1/2*f*x + 1/2*e) - 36*b*
c^2*d*tan(1/2*f*x + 1/2*e) - 36*a*c*d^2*tan(1/2*f*x + 1/2*e) - 72*b*c*d^2*tan(1/2*f*x + 1/2*e) - 24*a*d^3*tan(
1/2*f*x + 1/2*e) - 15*b*d^3*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^4)/f