3.153 \(\int \csc ^3(a+b x) \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=49 \[ \frac{3 \sec (a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b} \]

[Out]

(-3*ArcTanh[Cos[a + b*x]])/(2*b) + (3*Sec[a + b*x])/(2*b) - (Csc[a + b*x]^2*Sec[a + b*x])/(2*b)

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Rubi [A]  time = 0.0429918, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2622, 288, 321, 207} \[ \frac{3 \sec (a+b x)}{2 b}-\frac{3 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(-3*ArcTanh[Cos[a + b*x]])/(2*b) + (3*Sec[a + b*x])/(2*b) - (Csc[a + b*x]^2*Sec[a + b*x])/(2*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 ^3(a+b x) \sec ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (a+b x)\right )}{b}\\ &=-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{2 b}\\ &=\frac{3 \sec (a+b x)}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{2 b}\\ &=-\frac{3 \tanh ^{-1}(\cos (a+b x))}{2 b}+\frac{3 \sec (a+b x)}{2 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.248013, size = 143, normalized size = 2.92 \[ \frac{\csc ^4(a+b x) \left (-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (3 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-2\right )+2\right )}{2 b \left (\csc ^2\left (\frac{1}{2} (a+b x)\right )-\sec ^2\left (\frac{1}{2} (a+b x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sec[a + b*x]^2,x]

[Out]

(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Co
s[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))
/(2*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))

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Maple [A]  time = 0.017, size = 57, normalized size = 1.2 \begin{align*} -{\frac{1}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}\cos \left ( bx+a \right ) }}+{\frac{3}{2\,b\cos \left ( bx+a \right ) }}+{\frac{3\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^2/sin(b*x+a)^3,x)

[Out]

-1/2/b/sin(b*x+a)^2/cos(b*x+a)+3/2/b/cos(b*x+a)+3/2/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.970174, size = 82, normalized size = 1.67 \begin{align*} \frac{\frac{2 \,{\left (3 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )} - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/4*(2*(3*cos(b*x + a)^2 - 2)/(cos(b*x + a)^3 - cos(b*x + a)) - 3*log(cos(b*x + a) + 1) + 3*log(cos(b*x + a) -
1))/b

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Fricas [B]  time = 2.25569, size = 261, normalized size = 5.33 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{2} - 3 \,{\left (\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 ) + 3 \,{\left (\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 ) - 4}{4 \,{\left (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)^3,x, algorithm="fricas")

[Out]

1/4*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - co
s(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**2/sin(b*x+a)**3,x)

[Out]

Integral(sec(a + b*x)**2/sin(a + b*x)**3, x)

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Giac [B]  time = 1.20574, size = 189, normalized size = 3.86 \begin{align*} \frac{\frac{\frac{14 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1}{\frac{\cos \left (b x + a\right ) - 1}{\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{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 6 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^2/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*((14*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2) - (cos(b*x + a) - 1)/(cos(b*x + a) + 1
) + 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b