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

Optimal. Leaf size=107 \[ \frac{b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+a^4 x+\frac{4 a b^3 \tan (c+d x) \sec (c+d x)}{3 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

a^4*x + (2*a*b*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/d + (b^2*(17*a^2 + 2*b^2)*Tan[c + d*x])/(3*d) + (4*a*b^3*S
ec[c + d*x]*Tan[c + d*x])/(3*d) + (b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.115748, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3782, 4048, 3770, 3767, 8} \[ \frac{b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+a^4 x+\frac{4 a b^3 \tan (c+d x) \sec (c+d x)}{3 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4,x]

[Out]

a^4*x + (2*a*b*(2*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/d + (b^2*(17*a^2 + 2*b^2)*Tan[c + d*x])/(3*d) + (4*a*b^3*S
ec[c + d*x]*Tan[c + d*x])/(3*d) + (b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, 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 (a+b \sec (c+d x))^4 \, dx &=\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \sec (c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \sec (c+d x)+8 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \sec (c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^4 x+\frac{4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\left (2 a b \left (2 a^2+b^2\right )\right ) \int \sec (c+d x) \, dx+\frac{1}{3} \left (b^2 \left (17 a^2+2 b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=a^4 x+\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{\left (b^2 \left (17 a^2+2 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=a^4 x+\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \left (17 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac{4 a b^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac{b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.284188, size = 77, normalized size = 0.72 \[ \frac{6 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))+3 b^2 \tan (c+d x) \left (6 a^2+2 a b \sec (c+d x)+b^2\right )+3 a^4 d x+b^4 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4,x]

[Out]

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

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Maple [A]  time = 0.031, size = 135, normalized size = 1.3 \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}+4\,{\frac{{a}^{3}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{2\,{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{4}\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((a+b*sec(d*x+c))^4,x)

[Out]

a^4*x+1/d*a^4*c+4/d*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+6/d*a^2*b^2*tan(d*x+c)+2*a*b^3*sec(d*x+c)*tan(d*x+c)/d+2/d
*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*b^4*tan(d*x+c)+1/3/d*b^4*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.20451, size = 163, normalized size = 1.52 \begin{align*} a^{4} x + \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{4}}{3 \, d} - \frac{a b^{3}{\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 )}}{d} + \frac{4 \, a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{6 \, a^{2} b^{2} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70564, size = 339, normalized size = 3.17 \begin{align*} \frac{3 \, a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, a b^{3} \cos \left (d x + c\right ) + b^{4} + 2 \,{\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(c + d*x))**4, x)

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Giac [B]  time = 1.29683, size = 298, normalized size = 2.79 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{4} + 6 \,{\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \,{\left (2 \, a^{3} b + a b^{3}\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^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{4} \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}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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