3.155 \(\int \csc ^3(a+b x) \sec ^4(a+b x) \, dx\)

Optimal. Leaf size=66 \[ \frac{5 \sec ^3(a+b x)}{6 b}+\frac{5 \sec (a+b x)}{2 b}-\frac{5 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\csc ^2(a+b x) \sec ^3(a+b x)}{2 b} \]

[Out]

(-5*ArcTanh[Cos[a + b*x]])/(2*b) + (5*Sec[a + b*x])/(2*b) + (5*Sec[a + b*x]^3)/(6*b) - (Csc[a + b*x]^2*Sec[a +
b*x]^3)/(2*b)

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Rubi [A]  time = 0.0427421, antiderivative size = 66, 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, 302, 207} \[ \frac{5 \sec ^3(a+b x)}{6 b}+\frac{5 \sec (a+b x)}{2 b}-\frac{5 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{\csc ^2(a+b x) \sec ^3(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sec[a + b*x]^4,x]

[Out]

(-5*ArcTanh[Cos[a + b*x]])/(2*b) + (5*Sec[a + b*x])/(2*b) + (5*Sec[a + b*x]^3)/(6*b) - (Csc[a + b*x]^2*Sec[a +
b*x]^3)/(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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [B]  time = 0.421722, size = 205, normalized size = 3.11 \[ \frac{2 \csc ^8(a+b x) \left (-40 \cos (2 (a+b x))+13 \cos (3 (a+b x))-30 \cos (4 (a+b x))+13 \cos (5 (a+b x))+15 \cos (3 (a+b x)) \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )+15 \cos (5 (a+b x)) \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-15 \cos (3 (a+b x)) \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )-15 \cos (5 (a+b x)) \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (30 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )-30 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-26\right )+22\right )}{3 b \left (\csc ^2\left (\frac{1}{2} (a+b x)\right )-\sec ^2\left (\frac{1}{2} (a+b x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sec[a + b*x]^4,x]

[Out]

(2*Csc[a + b*x]^8*(22 - 40*Cos[2*(a + b*x)] + 13*Cos[3*(a + b*x)] - 30*Cos[4*(a + b*x)] + 13*Cos[5*(a + b*x)]
+ 15*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] + 15*Cos[5*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 15*Cos[3*(a + b*x)]*
Log[Sin[(a + b*x)/2]] - 15*Cos[5*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-26 - 30*Log[Cos[(a + b*x)/2
]] + 30*Log[Sin[(a + b*x)/2]])))/(3*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)

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Maple [A]  time = 0.023, size = 78, normalized size = 1.2 \begin{align*}{\frac{1}{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{3}}}-{\frac{5}{6\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}\cos \left ( bx+a \right ) }}+{\frac{5}{2\,b\cos \left ( bx+a \right ) }}+{\frac{5\,\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)^4/sin(b*x+a)^3,x)

[Out]

1/3/b/sin(b*x+a)^2/cos(b*x+a)^3-5/6/b/sin(b*x+a)^2/cos(b*x+a)+5/2/b/cos(b*x+a)+5/2/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.997612, size = 99, normalized size = 1.5 \begin{align*} \frac{\frac{2 \,{\left (15 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} - 2\right )}}{\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

1/12*(2*(15*cos(b*x + a)^4 - 10*cos(b*x + a)^2 - 2)/(cos(b*x + a)^5 - cos(b*x + a)^3) - 15*log(cos(b*x + a) +
1) + 15*log(cos(b*x + a) - 1))/b

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Fricas [A]  time = 2.32969, size = 301, normalized size = 4.56 \begin{align*} \frac{30 \, \cos \left (b x + a\right )^{4} - 20 \, \cos \left (b x + a\right )^{2} - 15 \,{\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\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} - \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - 4}{12 \,{\left (b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

1/12*(30*cos(b*x + a)^4 - 20*cos(b*x + a)^2 - 15*(cos(b*x + a)^5 - cos(b*x + a)^3)*log(1/2*cos(b*x + a) + 1/2)
+ 15*(cos(b*x + a)^5 - cos(b*x + a)^3)*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^5 - b*cos(b*x + a)^3
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\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)**4/sin(b*x+a)**3,x)

[Out]

Integral(sec(a + b*x)**4/sin(a + b*x)**3, x)

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Giac [B]  time = 1.17862, size = 220, normalized size = 3.33 \begin{align*} -\frac{\frac{3 \,{\left (\frac{10 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )}{\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 )}}{\cos \left (b x + a\right ) + 1} - \frac{16 \,{\left (\frac{12 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{9 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 7\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}} - 30 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4/sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/24*(3*(10*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) + 3*(cos(b*x + a
) - 1)/(cos(b*x + a) + 1) - 16*(12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 9*(cos(b*x + a) - 1)^2/(cos(b*x + a
) + 1)^2 + 7)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^3 - 30*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) +
1)))/b