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

Optimal. Leaf size=103 \[ \frac{2 a^2 (3 c+2 d) \tan (e+f x)}{3 f}+\frac{a^2 (3 c+2 d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{a^2 (3 c+2 d) \tan (e+f x) \sec (e+f x)}{6 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.117064, antiderivative size = 103, 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 = {4001, 3788, 3767, 8, 4046, 3770} \[ \frac{2 a^2 (3 c+2 d) \tan (e+f x)}{3 f}+\frac{a^2 (3 c+2 d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{a^2 (3 c+2 d) \tan (e+f x) \sec (e+f x)}{6 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4001

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[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

Mathematica [B]  time = 6.24232, size = 481, normalized size = 4.67 \[ \frac{a^2 \cos ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^2 (c+d \sec (e+f x)) \left (\frac{4 (6 c+5 d) \sin \left (\frac{f x}{2}\right )}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{4 (6 c+5 d) \sin \left (\frac{f x}{2}\right )}{\left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{(3 c+7 d) \cos \left (\frac{e}{2}\right )-(3 c+5 d) \sin \left (\frac{e}{2}\right )}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{(3 c+5 d) \sin \left (\frac{e}{2}\right )+(3 c+7 d) \cos \left (\frac{e}{2}\right )}{\left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-6 (3 c+2 d) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+6 (3 c+2 d) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{2 d \sin \left (\frac{f x}{2}\right )}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}+\frac{2 d \sin \left (\frac{f x}{2}\right )}{\left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}\right )}{48 f (c \cos (e+f x)+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[e + f*x]^3*Sec[(e + f*x)/2]^4*(1 + Sec[e + f*x])^2*(c + d*Sec[e + f*x])*(-6*(3*c + 2*d)*Log[Cos[(e +
f*x)/2] - Sin[(e + f*x)/2]] + 6*(3*c + 2*d)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (2*d*Sin[(f*x)/2])/((Co
s[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3) + ((3*c + 7*d)*Cos[e/2] - (3*c + 5*d)*Sin[e/2])/((
Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2) + (4*(6*c + 5*d)*Sin[(f*x)/2])/((Cos[e/2] - Sin[
e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])) + (2*d*Sin[(f*x)/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^3) - ((3*c + 7*d)*Cos[e/2] + (3*c + 5*d)*Sin[e/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^2) + (4*(6*c + 5*d)*Sin[(f*x)/2])/((Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]))))/(48*f*(d + c*Cos[e + f*x]))

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Maple [A]  time = 0.041, size = 141, normalized size = 1.4 \begin{align*}{\frac{3\,{a}^{2}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{5\,{a}^{2}d\tan \left ( fx+e \right ) }{3\,f}}+2\,{\frac{{a}^{2}c\tan \left ( fx+e \right ) }{f}}+{\frac{{a}^{2}d\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+{\frac{{a}^{2}d\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{{a}^{2}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}d\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+a*sec(f*x+e))^2*(c+d*sec(f*x+e)),x)

[Out]

3/2/f*a^2*c*ln(sec(f*x+e)+tan(f*x+e))+5/3/f*a^2*d*tan(f*x+e)+2/f*a^2*c*tan(f*x+e)+1/f*a^2*d*sec(f*x+e)*tan(f*x
+e)+1/f*a^2*d*ln(sec(f*x+e)+tan(f*x+e))+1/2*a^2*c*sec(f*x+e)*tan(f*x+e)/f+1/3/f*a^2*d*tan(f*x+e)*sec(f*x+e)^2

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Maxima [A]  time = 0.98352, size = 225, normalized size = 2.18 \begin{align*} \frac{4 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} d - 3 \, a^{2} c{\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 )} - 6 \, a^{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 )} + 12 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 24 \, a^{2} c \tan \left (f x + e\right ) + 12 \, a^{2} 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+a*sec(f*x+e))^2*(c+d*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.490226, size = 336, normalized size = 3.26 \begin{align*} \frac{3 \,{\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, a^{2} d + 2 \,{\left (6 \, a^{2} c + 5 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (a^{2} c + 2 \, a^{2} d\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+a*sec(f*x+e))^2*(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(3*(3*a^2*c + 2*a^2*d)*cos(f*x + e)^3*log(sin(f*x + e) + 1) - 3*(3*a^2*c + 2*a^2*d)*cos(f*x + e)^3*log(-s
in(f*x + e) + 1) + 2*(2*a^2*d + 2*(6*a^2*c + 5*a^2*d)*cos(f*x + e)^2 + 3*(a^2*c + 2*a^2*d)*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*} a^{2} \left (\int c \sec{\left (e + f x \right )}\, dx + \int 2 c \sec ^{2}{\left (e + f x \right )}\, dx + \int c \sec ^{3}{\left (e + f x \right )}\, dx + \int d \sec ^{2}{\left (e + f x \right )}\, dx + \int 2 d \sec ^{3}{\left (e + f x \right )}\, dx + \int d \sec ^{4}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(c*sec(e + f*x), x) + Integral(2*c*sec(e + f*x)**2, x) + Integral(c*sec(e + f*x)**3, x) + Integr
al(d*sec(e + f*x)**2, x) + Integral(2*d*sec(e + f*x)**3, x) + Integral(d*sec(e + f*x)**4, x))

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Giac [A]  time = 1.43254, size = 252, normalized size = 2.45 \begin{align*} \frac{3 \,{\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, a^{2} c + 2 \, a^{2} d\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (9 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 6 \, a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 24 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 16 \, a^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 18 \, a^{2} d \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+a*sec(f*x+e))^2*(c+d*sec(f*x+e)),x, algorithm="giac")

[Out]

1/6*(3*(3*a^2*c + 2*a^2*d)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*(3*a^2*c + 2*a^2*d)*log(abs(tan(1/2*f*x + 1/
2*e) - 1)) - 2*(9*a^2*c*tan(1/2*f*x + 1/2*e)^5 + 6*a^2*d*tan(1/2*f*x + 1/2*e)^5 - 24*a^2*c*tan(1/2*f*x + 1/2*e
)^3 - 16*a^2*d*tan(1/2*f*x + 1/2*e)^3 + 15*a^2*c*tan(1/2*f*x + 1/2*e) + 18*a^2*d*tan(1/2*f*x + 1/2*e))/(tan(1/
2*f*x + 1/2*e)^2 - 1)^3)/f