3.72 \(\int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=136 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{43 \tan (c+d x)}{21 a^4 d (\sec (c+d x)+1)}+\frac{11 \tan (c+d x)}{21 a^4 d (\sec (c+d x)+1)^2}-\frac{\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 \tan (c+d x) \sec ^2(c+d x)}{7 a d (a \sec (c+d x)+a)^3} \]

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) + (11*Tan[c + d*x])/(21*a^4*d*(1 + Sec[c + d*x])^2) - (43*Tan[c + d*x])/(21*a^4*
d*(1 + Sec[c + d*x])) - (Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Sec[c + d*x]^2*Tan[c +
d*x])/(7*a*d*(a + a*Sec[c + d*x])^3)

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Rubi [A]  time = 0.323221, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3816, 4019, 4008, 3998, 3770, 3794} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{43 \tan (c+d x)}{21 a^4 d (\sec (c+d x)+1)}+\frac{11 \tan (c+d x)}{21 a^4 d (\sec (c+d x)+1)^2}-\frac{\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac{2 \tan (c+d x) \sec ^2(c+d x)}{7 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) + (11*Tan[c + d*x])/(21*a^4*d*(1 + Sec[c + d*x])^2) - (43*Tan[c + d*x])/(21*a^4*
d*(1 + Sec[c + d*x])) - (Sec[c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Sec[c + d*x]^2*Tan[c +
d*x])/(7*a*d*(a + a*Sec[c + d*x])^3)

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 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3794

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

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{\sec ^3(c+d x) (3 a-7 a \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^2(c+d x) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^2(c+d x) \left (20 a^2-35 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{11 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^2(c+d x) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) \left (-110 a^3+105 a^3 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac{11 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^2(c+d x) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3}+\frac{\int \sec (c+d x) \, dx}{a^4}-\frac{43 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{21 a^3}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{11 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))^2}-\frac{\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \sec ^2(c+d x) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3}-\frac{43 \tan (c+d x)}{21 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.907558, size = 193, normalized size = 1.42 \[ -\frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (\sec \left (\frac{c}{2}\right ) \left (-434 \sin \left (c+\frac{d x}{2}\right )+525 \sin \left (c+\frac{3 d x}{2}\right )-147 \sin \left (2 c+\frac{3 d x}{2}\right )+203 \sin \left (2 c+\frac{5 d x}{2}\right )-21 \sin \left (3 c+\frac{5 d x}{2}\right )+32 \sin \left (3 c+\frac{7 d x}{2}\right )+686 \sin \left (\frac{d x}{2}\right )\right )+1344 \cos ^7\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{84 a^4 d (\sec (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^4,x]

[Out]

-(Cos[(c + d*x)/2]*Sec[c + d*x]^4*(1344*Cos[(c + d*x)/2]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*(686*Sin[(d*x)/2] - 434*Sin[c + (d*x)/2] + 525*Sin[c + (3*d*x)/2
] - 147*Sin[2*c + (3*d*x)/2] + 203*Sin[2*c + (5*d*x)/2] - 21*Sin[3*c + (5*d*x)/2] + 32*Sin[3*c + (7*d*x)/2])))
/(84*a^4*d*(1 + Sec[c + d*x])^4)

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Maple [A]  time = 0.039, size = 115, normalized size = 0.9 \begin{align*} -{\frac{1}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{11}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{15}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \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))^4,x)

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7-1/8/d/a^4*tan(1/2*d*x+1/2*c)^5-11/24/d/a^4*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*ta
n(1/2*d*x+1/2*c)-1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)

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Maxima [A]  time = 1.18645, size = 188, normalized size = 1.38 \begin{align*} -\frac{\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{168 \, 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))^4,x, algorithm="maxima")

[Out]

-1/168*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a
^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d

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Fricas [A]  time = 1.75158, size = 539, normalized size = 3.96 \begin{align*} \frac{21 \,{\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \,{\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (32 \, \cos \left (d x + c\right )^{3} + 107 \, \cos \left (d x + c\right )^{2} + 124 \, \cos \left (d x + c\right ) + 52\right )} \sin \left (d x + c\right )}{42 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\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))^4,x, algorithm="fricas")

[Out]

1/42*(21*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 2
1*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(32*c
os(d*x + c)^3 + 107*cos(d*x + c)^2 + 124*cos(d*x + c) + 52)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(
d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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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 ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx}{a^{4}} \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))**4,x)

[Out]

Integral(sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)/a*
*4

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Giac [A]  time = 1.36403, size = 149, normalized size = 1.1 \begin{align*} \frac{\frac{168 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{168 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{3 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 21 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 315 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{168 \, 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))^4,x, algorithm="giac")

[Out]

1/168*(168*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 168*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (3*a^24*tan(1
/2*d*x + 1/2*c)^7 + 21*a^24*tan(1/2*d*x + 1/2*c)^5 + 77*a^24*tan(1/2*d*x + 1/2*c)^3 + 315*a^24*tan(1/2*d*x + 1
/2*c))/a^28)/d