3.170 \(\int \csc ^4(a+b x) \sec ^5(a+b x) \, dx\)

Optimal. Leaf size=89 \[ -\frac{35 \csc ^3(a+b x)}{24 b}-\frac{35 \csc (a+b x)}{8 b}+\frac{35 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}+\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b} \]

[Out]

(35*ArcTanh[Sin[a + b*x]])/(8*b) - (35*Csc[a + b*x])/(8*b) - (35*Csc[a + b*x]^3)/(24*b) + (7*Csc[a + b*x]^3*Se
c[a + b*x]^2)/(8*b) + (Csc[a + b*x]^3*Sec[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0493638, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2621, 288, 302, 207} \[ -\frac{35 \csc ^3(a+b x)}{24 b}-\frac{35 \csc (a+b x)}{8 b}+\frac{35 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}+\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4*Sec[a + b*x]^5,x]

[Out]

(35*ArcTanh[Sin[a + b*x]])/(8*b) - (35*Csc[a + b*x])/(8*b) - (35*Csc[a + b*x]^3)/(24*b) + (7*Csc[a + b*x]^3*Se
c[a + b*x]^2)/(8*b) + (Csc[a + b*x]^3*Sec[a + b*x]^4)/(4*b)

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(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 ^4(a+b x) \sec ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^8}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac{7 \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=-\frac{35 \csc (a+b x)}{8 b}-\frac{35 \csc ^3(a+b x)}{24 b}+\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}-\frac{35 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac{35 \tanh ^{-1}(\sin (a+b x))}{8 b}-\frac{35 \csc (a+b x)}{8 b}-\frac{35 \csc ^3(a+b x)}{24 b}+\frac{7 \csc ^3(a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc ^3(a+b x) \sec ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [C]  time = 0.0139726, size = 31, normalized size = 0.35 \[ -\frac{\csc ^3(a+b x) \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};\sin ^2(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^4*Sec[a + b*x]^5,x]

[Out]

-(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[a + b*x]^2])/(3*b)

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Maple [A]  time = 0.026, size = 97, normalized size = 1.1 \begin{align*}{\frac{1}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3} \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}-{\frac{7}{12\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3} \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{35}{24\,b\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{35}{8\,b\sin \left ( bx+a \right ) }}+{\frac{35\,\ln \left ( \sec \left ( bx+a \right ) +\tan \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)^5/sin(b*x+a)^4,x)

[Out]

1/4/b/sin(b*x+a)^3/cos(b*x+a)^4-7/12/b/sin(b*x+a)^3/cos(b*x+a)^2+35/24/b/sin(b*x+a)/cos(b*x+a)^2-35/8/b/sin(b*
x+a)+35/8/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [A]  time = 0.996, size = 123, normalized size = 1.38 \begin{align*} -\frac{\frac{2 \,{\left (105 \, \sin \left (b x + a\right )^{6} - 175 \, \sin \left (b x + a\right )^{4} + 56 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{7} - 2 \, \sin \left (b x + a\right )^{5} + \sin \left (b x + a\right )^{3}} - 105 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/48*(2*(105*sin(b*x + a)^6 - 175*sin(b*x + a)^4 + 56*sin(b*x + a)^2 + 8)/(sin(b*x + a)^7 - 2*sin(b*x + a)^5
+ sin(b*x + a)^3) - 105*log(sin(b*x + a) + 1) + 105*log(sin(b*x + a) - 1))/b

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Fricas [A]  time = 2.33682, size = 375, normalized size = 4.21 \begin{align*} -\frac{210 \, \cos \left (b x + a\right )^{6} - 280 \, \cos \left (b x + a\right )^{4} - 105 \,{\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 105 \,{\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 42 \, \cos \left (b x + a\right )^{2} + 12}{48 \,{\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/48*(210*cos(b*x + a)^6 - 280*cos(b*x + a)^4 - 105*(cos(b*x + a)^6 - cos(b*x + a)^4)*log(sin(b*x + a) + 1)*s
in(b*x + a) + 105*(cos(b*x + a)^6 - cos(b*x + a)^4)*log(-sin(b*x + a) + 1)*sin(b*x + a) + 42*cos(b*x + a)^2 +
12)/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*sin(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)**5/sin(b*x+a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.21251, size = 115, normalized size = 1.29 \begin{align*} -\frac{\frac{6 \,{\left (11 \, \sin \left (b x + a\right )^{3} - 13 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac{16 \,{\left (9 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 105 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 105 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^5/sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/48*(6*(11*sin(b*x + a)^3 - 13*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 16*(9*sin(b*x + a)^2 + 1)/sin(b*x + a)
^3 - 105*log(abs(sin(b*x + a) + 1)) + 105*log(abs(sin(b*x + a) - 1)))/b