3.7 \(\int \csc ^7(a+b x) \, dx\)

Optimal. Leaf size=76 \[ -\frac {5 \tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}-\frac {5 \cot (a+b x) \csc ^3(a+b x)}{24 b}-\frac {5 \cot (a+b x) \csc (a+b x)}{16 b} \]

[Out]

-5/16*arctanh(cos(b*x+a))/b-5/16*cot(b*x+a)*csc(b*x+a)/b-5/24*cot(b*x+a)*csc(b*x+a)^3/b-1/6*cot(b*x+a)*csc(b*x
+a)^5/b

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Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3768, 3770} \[ -\frac {5 \tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}-\frac {5 \cot (a+b x) \csc ^3(a+b x)}{24 b}-\frac {5 \cot (a+b x) \csc (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^7,x]

[Out]

(-5*ArcTanh[Cos[a + b*x]])/(16*b) - (5*Cot[a + b*x]*Csc[a + b*x])/(16*b) - (5*Cot[a + b*x]*Csc[a + b*x]^3)/(24
*b) - (Cot[a + b*x]*Csc[a + b*x]^5)/(6*b)

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin {align*} \int \csc ^7(a+b x) \, dx &=-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}+\frac {5}{6} \int \csc ^5(a+b x) \, dx\\ &=-\frac {5 \cot (a+b x) \csc ^3(a+b x)}{24 b}-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}+\frac {5}{8} \int \csc ^3(a+b x) \, dx\\ &=-\frac {5 \cot (a+b x) \csc (a+b x)}{16 b}-\frac {5 \cot (a+b x) \csc ^3(a+b x)}{24 b}-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}+\frac {5}{16} \int \csc (a+b x) \, dx\\ &=-\frac {5 \tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {5 \cot (a+b x) \csc (a+b x)}{16 b}-\frac {5 \cot (a+b x) \csc ^3(a+b x)}{24 b}-\frac {\cot (a+b x) \csc ^5(a+b x)}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 151, normalized size = 1.99 \[ -\frac {\csc ^6\left (\frac {1}{2} (a+b x)\right )}{384 b}-\frac {\csc ^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {5 \csc ^2\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {\sec ^6\left (\frac {1}{2} (a+b x)\right )}{384 b}+\frac {\sec ^4\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {5 \sec ^2\left (\frac {1}{2} (a+b x)\right )}{64 b}+\frac {5 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{16 b}-\frac {5 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^7,x]

[Out]

(-5*Csc[(a + b*x)/2]^2)/(64*b) - Csc[(a + b*x)/2]^4/(64*b) - Csc[(a + b*x)/2]^6/(384*b) - (5*Log[Cos[(a + b*x)
/2]])/(16*b) + (5*Log[Sin[(a + b*x)/2]])/(16*b) + (5*Sec[(a + b*x)/2]^2)/(64*b) + Sec[(a + b*x)/2]^4/(64*b) +
Sec[(a + b*x)/2]^6/(384*b)

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fricas [B]  time = 0.76, size = 155, normalized size = 2.04 \[ \frac {30 \, \cos \left (b x + a\right )^{5} - 80 \, \cos \left (b x + a\right )^{3} - 15 \, {\left (\cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 66 \, \cos \left (b x + a\right )}{96 \, {\left (b \cos \left (b x + a\right )^{6} - 3 \, b \cos \left (b x + a\right )^{4} + 3 \, b \cos \left (b x + a\right )^{2} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^7,x, algorithm="fricas")

[Out]

1/96*(30*cos(b*x + a)^5 - 80*cos(b*x + a)^3 - 15*(cos(b*x + a)^6 - 3*cos(b*x + a)^4 + 3*cos(b*x + a)^2 - 1)*lo
g(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*x + a)^6 - 3*cos(b*x + a)^4 + 3*cos(b*x + a)^2 - 1)*log(-1/2*cos(b*x + a
) + 1/2) + 66*cos(b*x + a))/(b*cos(b*x + a)^6 - 3*b*cos(b*x + a)^4 + 3*b*cos(b*x + a)^2 - b)

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giac [B]  time = 0.21, size = 182, normalized size = 2.39 \[ -\frac {\frac {{\left (\frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {45 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {110 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{3}} + \frac {45 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - 60 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^7,x, algorithm="giac")

[Out]

-1/384*((9*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 45*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 110*(cos(b*x
+ a) - 1)^3/(cos(b*x + a) + 1)^3 - 1)*(cos(b*x + a) + 1)^3/(cos(b*x + a) - 1)^3 + 45*(cos(b*x + a) - 1)/(cos(
b*x + a) + 1) - 9*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + (cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 60*l
og(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

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maple [A]  time = 0.78, size = 78, normalized size = 1.03 \[ -\frac {\cot \left (b x +a \right ) \left (\csc ^{5}\left (b x +a \right )\right )}{6 b}-\frac {5 \cot \left (b x +a \right ) \left (\csc ^{3}\left (b x +a \right )\right )}{24 b}-\frac {5 \cot \left (b x +a \right ) \csc \left (b x +a \right )}{16 b}+\frac {5 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^7,x)

[Out]

-1/6*cot(b*x+a)*csc(b*x+a)^5/b-5/24*cot(b*x+a)*csc(b*x+a)^3/b-5/16*cot(b*x+a)*csc(b*x+a)/b+5/16/b*ln(csc(b*x+a
)-cot(b*x+a))

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maxima [A]  time = 0.34, size = 91, normalized size = 1.20 \[ \frac {\frac {2 \, {\left (15 \, \cos \left (b x + a\right )^{5} - 40 \, \cos \left (b x + a\right )^{3} + 33 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} + 3 \, \cos \left (b x + a\right )^{2} - 1} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^7,x, algorithm="maxima")

[Out]

1/96*(2*(15*cos(b*x + a)^5 - 40*cos(b*x + a)^3 + 33*cos(b*x + a))/(cos(b*x + a)^6 - 3*cos(b*x + a)^4 + 3*cos(b
*x + a)^2 - 1) - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x + a) - 1))/b

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mupad [B]  time = 0.15, size = 78, normalized size = 1.03 \[ \frac {\frac {5\,{\cos \left (a+b\,x\right )}^5}{16}-\frac {5\,{\cos \left (a+b\,x\right )}^3}{6}+\frac {11\,\cos \left (a+b\,x\right )}{16}}{b\,\left ({\cos \left (a+b\,x\right )}^6-3\,{\cos \left (a+b\,x\right )}^4+3\,{\cos \left (a+b\,x\right )}^2-1\right )}-\frac {5\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{16\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(a + b*x)^7,x)

[Out]

((11*cos(a + b*x))/16 - (5*cos(a + b*x)^3)/6 + (5*cos(a + b*x)^5)/16)/(b*(3*cos(a + b*x)^2 - 3*cos(a + b*x)^4
+ cos(a + b*x)^6 - 1)) - (5*atanh(cos(a + b*x)))/(16*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc ^{7}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**7,x)

[Out]

Integral(csc(a + b*x)**7, x)

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