3.181 \(\int \csc ^5(a+b x) \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=70 \[ \frac{15 \sec (a+b x)}{8 b}-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}-\frac{5 \csc ^2(a+b x) \sec (a+b x)}{8 b} \]

[Out]

(-15*ArcTanh[Cos[a + b*x]])/(8*b) + (15*Sec[a + b*x])/(8*b) - (5*Csc[a + b*x]^2*Sec[a + b*x])/(8*b) - (Csc[a +
b*x]^4*Sec[a + b*x])/(4*b)

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Rubi [A]  time = 0.0428248, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2622, 288, 321, 207} \[ \frac{15 \sec (a+b x)}{8 b}-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b}-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}-\frac{5 \csc ^2(a+b x) \sec (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^5*Sec[a + b*x]^2,x]

[Out]

(-15*ArcTanh[Cos[a + b*x]])/(8*b) + (15*Sec[a + b*x])/(8*b) - (5*Csc[a + b*x]^2*Sec[a + b*x])/(8*b) - (Csc[a +
b*x]^4*Sec[a + b*x])/(4*b)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \csc ^5(a+b x) \sec ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (a+b x)\right )}{4 b}\\ &=-\frac{5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{8 b}\\ &=\frac{15 \sec (a+b x)}{8 b}-\frac{5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{8 b}\\ &=-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b}+\frac{15 \sec (a+b x)}{8 b}-\frac{5 \csc ^2(a+b x) \sec (a+b x)}{8 b}-\frac{\csc ^4(a+b x) \sec (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 4.55911, size = 129, normalized size = 1.84 \[ -\frac{\csc ^4\left (\frac{1}{2} (a+b x)\right )+14 \csc ^2\left (\frac{1}{2} (a+b x)\right )+\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right ) \left (-14 \tan ^2\left (\frac{1}{2} (a+b x)\right )+\cos (a+b x) \left (\sec ^4\left (\frac{1}{2} (a+b x)\right )-8 \left (-15 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )+15 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )+8\right )\right )+78\right )}{\tan ^2\left (\frac{1}{2} (a+b x)\right )-1}}{64 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^5*Sec[a + b*x]^2,x]

[Out]

-(14*Csc[(a + b*x)/2]^2 + Csc[(a + b*x)/2]^4 + (Sec[(a + b*x)/2]^2*(78 + Cos[a + b*x]*(-8*(8 + 15*Log[Cos[(a +
b*x)/2]] - 15*Log[Sin[(a + b*x)/2]]) + Sec[(a + b*x)/2]^4) - 14*Tan[(a + b*x)/2]^2))/(-1 + Tan[(a + b*x)/2]^2
))/(64*b)

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Maple [A]  time = 0.022, size = 78, normalized size = 1.1 \begin{align*} -{\frac{1}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}\cos \left ( bx+a \right ) }}-{\frac{5}{8\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}\cos \left ( bx+a \right ) }}+{\frac{15}{8\,b\cos \left ( bx+a \right ) }}+{\frac{15\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2/sin(b*x+a)^5,x)

[Out]

-1/4/b/sin(b*x+a)^4/cos(b*x+a)-5/8/b/sin(b*x+a)^2/cos(b*x+a)+15/8/b/cos(b*x+a)+15/8/b*ln(csc(b*x+a)-cot(b*x+a)
)

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Maxima [A]  time = 0.968699, size = 107, normalized size = 1.53 \begin{align*} \frac{\frac{2 \,{\left (15 \, \cos \left (b x + a\right )^{4} - 25 \, \cos \left (b x + a\right )^{2} + 8\right )}}{\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/16*(2*(15*cos(b*x + a)^4 - 25*cos(b*x + a)^2 + 8)/(cos(b*x + a)^5 - 2*cos(b*x + a)^3 + cos(b*x + a)) - 15*lo
g(cos(b*x + a) + 1) + 15*log(cos(b*x + a) - 1))/b

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Fricas [B]  time = 1.98859, size = 374, normalized size = 5.34 \begin{align*} \frac{30 \, \cos \left (b x + a\right )^{4} - 50 \, \cos \left (b x + a\right )^{2} - 15 \,{\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 15 \,{\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 16}{16 \,{\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/16*(30*cos(b*x + a)^4 - 50*cos(b*x + a)^2 - 15*(cos(b*x + a)^5 - 2*cos(b*x + a)^3 + cos(b*x + a))*log(1/2*co
s(b*x + a) + 1/2) + 15*(cos(b*x + a)^5 - 2*cos(b*x + a)^3 + cos(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) + 16)/(
b*cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2/sin(b*x+a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.18195, size = 220, normalized size = 3.14 \begin{align*} \frac{\frac{{\left (\frac{16 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{90 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac{16 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{128}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1} + 60 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^5,x, algorithm="giac")

[Out]

1/64*((16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 90*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 1)*(cos(b*x +
a) + 1)^2/(cos(b*x + a) - 1)^2 - 16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a
) + 1)^2 + 128/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1) + 60*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) +
1)))/b