3.468 \(\int \sec (c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=99 \[ \frac{2 b \left (4 a^2+b^2\right ) \tan (c+d x)}{3 d}+\frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a b^2 \tan (c+d x) \sec (c+d x)}{6 d}+\frac{b \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

(a*(2*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*b*(4*a^2 + b^2)*Tan[c + d*x])/(3*d) + (5*a*b^2*Sec[c + d*
x]*Tan[c + d*x])/(6*d) + (b*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.131079, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3830, 3997, 3787, 3770, 3767, 8} \[ \frac{2 b \left (4 a^2+b^2\right ) \tan (c+d x)}{3 d}+\frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a b^2 \tan (c+d x) \sec (c+d x)}{6 d}+\frac{b \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*a^2 + 3*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (2*b*(4*a^2 + b^2)*Tan[c + d*x])/(3*d) + (5*a*b^2*Sec[c + d*
x]*Tan[c + d*x])/(6*d) + (b*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

Rule 3830

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

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 (c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{b (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) \left (3 a^2+2 b^2+5 a b \sec (c+d x)\right ) \, dx\\ &=\frac{5 a b^2 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{6} \int \sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )+4 b \left (4 a^2+b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{5 a b^2 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \left (2 b \left (4 a^2+b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{2} \left (a \left (2 a^2+3 b^2\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a b^2 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{\left (2 b \left (4 a^2+b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a \left (2 a^2+3 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 b \left (4 a^2+b^2\right ) \tan (c+d x)}{3 d}+\frac{5 a b^2 \sec (c+d x) \tan (c+d x)}{6 d}+\frac{b (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.240821, size = 70, normalized size = 0.71 \[ \frac{\left (6 a^3+9 a b^2\right ) \tanh ^{-1}(\sin (c+d x))+b \tan (c+d x) \left (18 a^2+9 a b \sec (c+d x)+2 b^2 \tan ^2(c+d x)+6 b^2\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^3 + 9*a*b^2)*ArcTanh[Sin[c + d*x]] + b*Tan[c + d*x]*(18*a^2 + 6*b^2 + 9*a*b*Sec[c + d*x] + 2*b^2*Tan[c +
d*x]^2))/(6*d)

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Maple [A]  time = 0.029, size = 118, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{3\,a{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+b*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.08217, size = 143, normalized size = 1.44 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{3} - 9 \, a b^{2}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, a^{2} b \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*b^3 - 9*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c
) + 1) + log(sin(d*x + c) - 1)) + 12*a^3*log(sec(d*x + c) + tan(d*x + c)) + 36*a^2*b*tan(d*x + c))/d

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Fricas [A]  time = 1.73864, size = 309, normalized size = 3.12 \begin{align*} \frac{3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (9 \, a b^{2} \cos \left (d x + c\right ) + 2 \, b^{3} + 2 \,{\left (9 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(3*(2*a^3 + 3*a*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*a^3 + 3*a*b^2)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 2*(9*a*b^2*cos(d*x + c) + 2*b^3 + 2*(9*a^2*b + 2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x
+ c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{3} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))**3,x)

[Out]

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

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Giac [B]  time = 1.38141, size = 277, normalized size = 2.8 \begin{align*} \frac{3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*a^3 + 3*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) - 2*(18*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 9*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*b^3*tan(1/2*d*x + 1/2*c)^5 -
36*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 4*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b*tan(1/2*d*x + 1/2*c) + 9*a*b^2*tan(1
/2*d*x + 1/2*c) + 6*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d