3.62 \(\int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=128 \[ \frac{9 \tan (c+d x)}{5 a^3 d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{\tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{3 \tan (c+d x) \sec ^2(c+d x)}{5 a d (a \sec (c+d x)+a)^2} \]

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(a^3*d) + (9*Tan[c + d*x])/(5*a^3*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Se
c[c + d*x])^3) - (3*Sec[c + d*x]^2*Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^2) + (3*Tan[c + d*x])/(d*(a^3 + a
^3*Sec[c + d*x]))

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Rubi [A]  time = 0.264708, antiderivative size = 128, 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 = {3816, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac{9 \tan (c+d x)}{5 a^3 d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{\tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{3 \tan (c+d x) \sec ^2(c+d x)}{5 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(a^3*d) + (9*Tan[c + d*x])/(5*a^3*d) - (Sec[c + d*x]^3*Tan[c + d*x])/(5*d*(a + a*Se
c[c + d*x])^3) - (3*Sec[c + d*x]^2*Tan[c + d*x])/(5*a*d*(a + a*Sec[c + d*x])^2) + (3*Tan[c + d*x])/(d*(a^3 + a
^3*Sec[c + d*x]))

Rule 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

Rule 4019

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

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-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 \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^3(c+d x) (3 a-6 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) \left (18 a^2-27 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec (c+d x) \left (-45 a^3+27 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{9 \int \sec ^2(c+d x) \, dx}{5 a^3}-\frac{3 \int \sec (c+d x) \, dx}{a^3}\\ &=-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{9 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 a^3 d}\\ &=-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{9 \tan (c+d x)}{5 a^3 d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.25505, size = 294, normalized size = 2.3 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (8 \tan \left (\frac{c}{2}\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+20 \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+76 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right )+8 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{5 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(Sec[c/2]*Sin[(d*x)/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 76*Co
s[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3
*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*
x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + Cos[(c + d*x)/2]*Tan[c/2] + 8*Cos[(c + d*x
)/2]^3*Tan[c/2]))/(5*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.039, size = 139, normalized size = 1.1 \begin{align*}{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}}-{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x)

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5+1/2/d/a^3*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*tan(1/2*d*x+1/2*c)-1/d/a^3/(tan(1/2*
d*x+1/2*c)+1)-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^3/(tan(1/2*d*x+1/2*c)-1)+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.19071, size = 223, normalized size = 1.74 \begin{align*} \frac{\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/20*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/
(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*lo
g(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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Fricas [A]  time = 1.7734, size = 506, normalized size = 3.95 \begin{align*} -\frac{15 \,{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (24 \, \cos \left (d x + c\right )^{3} + 57 \, \cos \left (d x + c\right )^{2} + 39 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{10 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/10*(15*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*log(sin(d*x + c) + 1) - 15*(co
s(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(24*cos(d*x + c)
^3 + 57*cos(d*x + c)^2 + 39*cos(d*x + c) + 5)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3
*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.4012, size = 165, normalized size = 1.29 \begin{align*} -\frac{\frac{60 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac{a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/20*(60*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 40*tan(1/2*d*x
+ 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (a^12*tan(1/2*d*x + 1/2*c)^5 + 10*a^12*tan(1/2*d*x + 1/2*c)^3 +
85*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d