3.467 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=130 \[ \frac{a \left (a^2+4 b^2\right ) \tan (c+d x)}{2 d}+\frac{3 b \left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{a \tan (c+d x) (a+b \sec (c+d x))^2}{4 d} \]

[Out]

(3*b*(4*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(a^2 + 4*b^2)*Tan[c + d*x])/(2*d) + (b*(2*a^2 + 3*b^2)*Se
c[c + d*x]*Tan[c + d*x])/(8*d) + (a*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(4*d) + ((a + b*Sec[c + d*x])^3*Tan[c
+ d*x])/(4*d)

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Rubi [A]  time = 0.198026, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3835, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{a \left (a^2+4 b^2\right ) \tan (c+d x)}{2 d}+\frac{3 b \left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{a \tan (c+d x) (a+b \sec (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3,x]

[Out]

(3*b*(4*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(a^2 + 4*b^2)*Tan[c + d*x])/(2*d) + (b*(2*a^2 + 3*b^2)*Se
c[c + d*x]*Tan[c + d*x])/(8*d) + (a*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(4*d) + ((a + b*Sec[c + d*x])^3*Tan[c
+ d*x])/(4*d)

Rule 3835

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

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 ^2(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{3}{4} \int \sec (c+d x) (b+a \sec (c+d x)) (a+b \sec (c+d x))^2 \, dx\\ &=\frac{a (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int \sec (c+d x) (a+b \sec (c+d x)) \left (5 a b+\left (2 a^2+3 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (2 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{8} \int \sec (c+d x) \left (3 b \left (4 a^2+b^2\right )+4 a \left (a^2+4 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b \left (2 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{8} \left (3 b \left (4 a^2+b^2\right )\right ) \int \sec (c+d x) \, dx+\frac{1}{2} \left (a \left (a^2+4 b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{3 b \left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \left (2 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a \left (a^2+4 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=\frac{3 b \left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \left (a^2+4 b^2\right ) \tan (c+d x)}{2 d}+\frac{b \left (2 a^2+3 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac{(a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.444184, size = 90, normalized size = 0.69 \[ \frac{3 b \left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 a \left (a^2+b^2 \tan ^2(c+d x)+3 b^2\right )+3 b \left (4 a^2+b^2\right ) \sec (c+d x)+2 b^3 \sec ^3(c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sec[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.029, size = 160, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{a{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.12123, size = 213, normalized size = 1.64 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{2} - b^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} b{\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 )} + 16 \, a^{3} \tan \left (d x + c\right )}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^2 - b^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4
- 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*a^2*b*(2*sin(d*x + c)/(sin(d
*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 16*a^3*tan(d*x + c))/d

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Fricas [A]  time = 1.70749, size = 348, normalized size = 2.68 \begin{align*} \frac{3 \,{\left (4 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \, a b^{2} \cos \left (d x + c\right ) + 8 \,{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, b^{3} + 3 \,{\left (4 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(3*(4*a^2*b + b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2*b + b^3)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(8*a*b^2*cos(d*x + c) + 8*(a^3 + 2*a*b^2)*cos(d*x + c)^3 + 2*b^3 + 3*(4*a^2*b + b^3)*cos(d*x + c)
^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*sec(c + d*x)**2, x)

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Giac [B]  time = 1.34871, size = 446, normalized size = 3.43 \begin{align*} \frac{3 \,{\left (4 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (4 \, a^{2} b + b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 5 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, 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 )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(3*(4*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2*b + b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) - 2*(8*a^3*tan(1/2*d*x + 1/2*c)^7 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 5*b
^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 40*a*b^2*tan(1/2
*d*x + 1/2*c)^5 - 3*b^3*tan(1/2*d*x + 1/2*c)^5 + 24*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*a^2*b*tan(1/2*d*x + 1/2*c)
^3 + 40*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 3*b^3*tan(1/2*d*x + 1/2*c)^3 - 8*a^3*tan(1/2*d*x + 1/2*c) - 12*a^2*b*ta
n(1/2*d*x + 1/2*c) - 24*a*b^2*tan(1/2*d*x + 1/2*c) - 5*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^
4)/d