3.486 \(\int (a+b \sec (c+d x))^5 \, dx\)

Optimal. Leaf size=158 \[ \frac{a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac{b \left (40 a^2 b^2+40 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (58 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^5 x+\frac{11 a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

[Out]

a^5*x + (b*(40*a^4 + 40*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b^2*(53*a^2 + 20*b^2)*Tan[c + d*x])
/(6*d) + (b^3*(58*a^2 + 9*b^2)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (11*a*b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*
x])/(12*d) + (b^2*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.236012, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3782, 4056, 4048, 3770, 3767, 8} \[ \frac{a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac{b \left (40 a^2 b^2+40 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (58 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+a^5 x+\frac{11 a b^2 \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{b^2 \tan (c+d x) (a+b \sec (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^5,x]

[Out]

a^5*x + (b*(40*a^4 + 40*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b^2*(53*a^2 + 20*b^2)*Tan[c + d*x])
/(6*d) + (b^3*(58*a^2 + 9*b^2)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (11*a*b^2*(a + b*Sec[c + d*x])^2*Tan[c + d*
x])/(12*d) + (b^2*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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 4056

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

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))^5 \, dx &=\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \sec (c+d x))^2 \left (4 a^3+3 b \left (4 a^2+b^2\right ) \sec (c+d x)+11 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \sec (c+d x)) \left (12 a^4+a b \left (48 a^2+31 b^2\right ) \sec (c+d x)+b^2 \left (58 a^2+9 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (24 a^5+3 b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \sec (c+d x)+4 a b^2 \left (53 a^2+20 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=a^5 x+\frac{b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (a b^2 \left (53 a^2+20 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (b \left (40 a^4+40 a^2 b^2+3 b^4\right )\right ) \int \sec (c+d x) \, dx\\ &=a^5 x+\frac{b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a b^2 \left (53 a^2+20 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=a^5 x+\frac{b \left (40 a^4+40 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a b^2 \left (53 a^2+20 b^2\right ) \tan (c+d x)}{6 d}+\frac{b^3 \left (58 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{11 a b^2 (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b^2 (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.565435, size = 114, normalized size = 0.72 \[ \frac{3 b \left (40 a^2 b^2+40 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))+3 b^2 \tan (c+d x) \left (b \left (40 a^2+3 b^2\right ) \sec (c+d x)+40 a \left (2 a^2+b^2\right )+2 b^3 \sec ^3(c+d x)\right )+24 a^5 d x+40 a b^4 \tan ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^5,x]

[Out]

(24*a^5*d*x + 3*b*(40*a^4 + 40*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]] + 3*b^2*(40*a*(2*a^2 + b^2) + b*(40*a^2
+ 3*b^2)*Sec[c + d*x] + 2*b^3*Sec[c + d*x]^3)*Tan[c + d*x] + 40*a*b^4*Tan[c + d*x]^3)/(24*d)

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Maple [A]  time = 0.042, size = 205, normalized size = 1.3 \begin{align*}{a}^{5}x+{\frac{{a}^{5}c}{d}}+5\,{\frac{{a}^{4}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+10\,{\frac{{a}^{3}{b}^{2}\tan \left ( dx+c \right ) }{d}}+5\,{\frac{{a}^{2}{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+5\,{\frac{{a}^{2}{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{10\,a{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{5\,a{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{5}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{5}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{5}\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((a+b*sec(d*x+c))^5,x)

[Out]

a^5*x+1/d*a^5*c+5/d*a^4*b*ln(sec(d*x+c)+tan(d*x+c))+10/d*a^3*b^2*tan(d*x+c)+5/d*a^2*b^3*sec(d*x+c)*tan(d*x+c)+
5/d*a^2*b^3*ln(sec(d*x+c)+tan(d*x+c))+10/3/d*a*b^4*tan(d*x+c)+5/3/d*a*b^4*tan(d*x+c)*sec(d*x+c)^2+1/4/d*b^5*ta
n(d*x+c)*sec(d*x+c)^3+3/8/d*b^5*sec(d*x+c)*tan(d*x+c)+3/8/d*b^5*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.21271, size = 267, normalized size = 1.69 \begin{align*} a^{5} x + \frac{5 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{4}}{3 \, d} - \frac{b^{5}{\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 )}}{16 \, d} - \frac{5 \, a^{2} 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 )}}{2 \, d} + \frac{5 \, a^{4} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{10 \, a^{3} 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))^5,x, algorithm="maxima")

[Out]

a^5*x + 5/3*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^4/d - 1/16*b^5*(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 - 5/2*a^2*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 + 5*a^4*b*log(sec(d*x + c) +
tan(d*x + c))/d + 10*a^3*b^2*tan(d*x + c)/d

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Fricas [A]  time = 1.80867, size = 444, normalized size = 2.81 \begin{align*} \frac{48 \, a^{5} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (40 \, a b^{4} \cos \left (d x + c\right ) + 6 \, b^{5} + 80 \,{\left (3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\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((a+b*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

1/48*(48*a^5*d*x*cos(d*x + c)^4 + 3*(40*a^4*b + 40*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(
40*a^4*b + 40*a^2*b^3 + 3*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(40*a*b^4*cos(d*x + c) + 6*b^5 + 80*(
3*a^3*b^2 + a*b^4)*cos(d*x + c)^3 + 3*(40*a^2*b^3 + 3*b^5)*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 )^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**5,x)

[Out]

Integral((a + b*sec(c + d*x))**5, x)

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Giac [B]  time = 1.18383, size = 513, normalized size = 3.25 \begin{align*} \frac{24 \,{\left (d x + c\right )} a^{5} + 3 \,{\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (40 \, a^{4} b + 40 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (240 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 120 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 720 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 120 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 200 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 720 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 200 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 240 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 120 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 120 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, b^{5} \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((a+b*sec(d*x+c))^5,x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*a^5 + 3*(40*a^4*b + 40*a^2*b^3 + 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(40*a^4*b +
40*a^2*b^3 + 3*b^5)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(240*a^3*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*a^2*b^3*t
an(1/2*d*x + 1/2*c)^7 + 120*a*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*b^5*tan(1/2*d*x + 1/2*c)^7 - 720*a^3*b^2*tan(1/2
*d*x + 1/2*c)^5 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 200*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 9*b^5*tan(1/2*d*x +
1/2*c)^5 + 720*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 200*a*b^4*tan(1/2*d*x + 1
/2*c)^3 - 9*b^5*tan(1/2*d*x + 1/2*c)^3 - 240*a^3*b^2*tan(1/2*d*x + 1/2*c) - 120*a^2*b^3*tan(1/2*d*x + 1/2*c) -
120*a*b^4*tan(1/2*d*x + 1/2*c) - 15*b^5*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d