3.446 \(\int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=85 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.0665066, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3787, 3767, 3768, 3770} \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x]),x]

[Out]

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

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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

Mathematica [A]  time = 0.287118, size = 76, normalized size = 0.89 \[ \frac{a \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.023, size = 92, normalized size = 1.1 \begin{align*}{\frac{2\,a\tan \left ( dx+c \right ) }{3\,d}}+{\frac{a\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{b \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,b\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)^4*(a+b*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.20689, size = 128, normalized size = 1.51 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a - 3 \, b{\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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a - 3*b*(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)))/d

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Fricas [A]  time = 1.82608, size = 266, normalized size = 3.13 \begin{align*} \frac{9 \, b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \, a \cos \left (d x + c\right )^{3} + 9 \, b \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + 6 \, b\right )} \sin \left (d x + c\right )}{48 \, 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)^4*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(9*b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*b*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*a*cos(d*x +
c)^3 + 9*b*cos(d*x + c)^2 + 8*a*cos(d*x + c) + 6*b)*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 ) \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*sec(c + d*x)**4, x)

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Giac [B]  time = 1.3283, size = 221, normalized size = 2.6 \begin{align*} \frac{9 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 40 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, b \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}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(9*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*a*tan(1/2*d*x +
1/2*c)^7 - 15*b*tan(1/2*d*x + 1/2*c)^7 - 40*a*tan(1/2*d*x + 1/2*c)^5 - 9*b*tan(1/2*d*x + 1/2*c)^5 + 40*a*tan(1
/2*d*x + 1/2*c)^3 - 9*b*tan(1/2*d*x + 1/2*c)^3 - 24*a*tan(1/2*d*x + 1/2*c) - 15*b*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 - 1)^4)/d