3.167 \(\int \sec (e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=91 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{3 b (2 a+b) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{b \tan (e+f x) \sec ^3(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{4 f} \]

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sin[e + f*x]])/(8*f) + (3*b*(2*a + b)*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (b*S
ec[e + f*x]^3*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x])/(4*f)

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Rubi [A]  time = 0.0739133, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4147, 413, 385, 206} \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{3 b (2 a+b) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{b \tan (e+f x) \sec ^3(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sin[e + f*x]])/(8*f) + (3*b*(2*a + b)*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (b*S
ec[e + f*x]^3*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x])/(4*f)

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 385

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

Rule 206

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

Rubi steps

\begin{align*} \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-a x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f}-\frac{\operatorname{Subst}\left (\int \frac{-(a+b) (4 a+3 b)+a (4 a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 f}\\ &=\frac{3 b (2 a+b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{8 f}\\ &=\frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{3 b (2 a+b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.13585, size = 63, normalized size = 0.69 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}(\sin (e+f x))+b \tan (e+f x) \sec (e+f x) \left (8 a+2 b \sec ^2(e+f x)+3 b\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sin[e + f*x]] + b*Sec[e + f*x]*(8*a + 3*b + 2*b*Sec[e + f*x]^2)*Tan[e + f*x])
/(8*f)

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Maple [A]  time = 0.037, size = 125, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{ab\tan \left ( fx+e \right ) \sec \left ( fx+e \right ) }{f}}+{\frac{ab\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{{b}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{3\,{b}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{b}^{2}\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)^2)^2,x)

[Out]

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

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Maxima [A]  time = 1.01531, size = 161, normalized size = 1.77 \begin{align*} \frac{{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left ({\left (8 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )^{3} -{\left (8 \, a b + 5 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{16 \, 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)^2)^2,x, algorithm="maxima")

[Out]

1/16*((8*a^2 + 8*a*b + 3*b^2)*log(sin(f*x + e) + 1) - (8*a^2 + 8*a*b + 3*b^2)*log(sin(f*x + e) - 1) - 2*((8*a*
b + 3*b^2)*sin(f*x + e)^3 - (8*a*b + 5*b^2)*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.51094, size = 284, normalized size = 3.12 \begin{align*} \frac{{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left ({\left (8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sin \left (f x + e\right )}{16 \, 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)^2)^2,x, algorithm="fricas")

[Out]

1/16*((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4*log(sin(f*x + e) + 1) - (8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4*lo
g(-sin(f*x + e) + 1) + 2*((8*a*b + 3*b^2)*cos(f*x + e)^2 + 2*b^2)*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 ^{2}{\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)**2)**2,x)

[Out]

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

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Giac [A]  time = 1.21823, size = 171, normalized size = 1.88 \begin{align*} \frac{{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - \frac{2 \,{\left (8 \, a b \sin \left (f x + e\right )^{3} + 3 \, b^{2} \sin \left (f x + e\right )^{3} - 8 \, a b \sin \left (f x + e\right ) - 5 \, b^{2} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2}}}{16 \, 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)^2)^2,x, algorithm="giac")

[Out]

1/16*((8*a^2 + 8*a*b + 3*b^2)*log(sin(f*x + e) + 1) - (8*a^2 + 8*a*b + 3*b^2)*log(-sin(f*x + e) + 1) - 2*(8*a*
b*sin(f*x + e)^3 + 3*b^2*sin(f*x + e)^3 - 8*a*b*sin(f*x + e) - 5*b^2*sin(f*x + e))/(sin(f*x + e)^2 - 1)^2)/f