3.166 \(\int \csc ^5(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=100 \[ -\frac {3 a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {b \cot ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]

[Out]

-3/8*a*arctanh(cos(d*x+c))/d-b*cot(d*x+c)^2/d-1/4*b*cot(d*x+c)^4/d-3/8*a*cot(d*x+c)*csc(d*x+c)/d-1/4*a*cot(d*x
+c)*csc(d*x+c)^3/d+b*ln(tan(d*x+c))/d

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Rubi [A]  time = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3872, 2834, 2620, 266, 43, 3768, 3770} \[ -\frac {3 a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3 a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {b \cot ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + b*Sec[c + d*x]),x]

[Out]

(-3*a*ArcTanh[Cos[c + d*x]])/(8*d) - (b*Cot[c + d*x]^2)/d - (b*Cot[c + d*x]^4)/(4*d) - (3*a*Cot[c + d*x]*Csc[c
+ d*x])/(8*d) - (a*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d) + (b*Log[Tan[c + d*x]])/d

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2620

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

Rule 2834

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

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]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 164, normalized size = 1.64 \[ -\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {a \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {3 a \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {b \left (\csc ^4(c+d x)+2 \csc ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5*(a + b*Sec[c + d*x]),x]

[Out]

(-3*a*Csc[(c + d*x)/2]^2)/(32*d) - (a*Csc[(c + d*x)/2]^4)/(64*d) - (3*a*Log[Cos[(c + d*x)/2]])/(8*d) + (3*a*Lo
g[Sin[(c + d*x)/2]])/(8*d) - (b*(2*Csc[c + d*x]^2 + Csc[c + d*x]^4 + 4*Log[Cos[c + d*x]] - 4*Log[Sin[c + d*x]]
))/(4*d) + (3*a*Sec[(c + d*x)/2]^2)/(32*d) + (a*Sec[(c + d*x)/2]^4)/(64*d)

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fricas [B]  time = 0.51, size = 201, normalized size = 2.01 \[ \frac {6 \, a \cos \left (d x + c\right )^{3} + 8 \, b \cos \left (d x + c\right )^{2} - 10 \, a \cos \left (d x + c\right ) - 16 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left ({\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{2} + 3 \, a - 8 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{2} + 3 \, a + 8 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, b}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(6*a*cos(d*x + c)^3 + 8*b*cos(d*x + c)^2 - 10*a*cos(d*x + c) - 16*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2
+ b)*log(-cos(d*x + c)) - ((3*a - 8*b)*cos(d*x + c)^4 - 2*(3*a - 8*b)*cos(d*x + c)^2 + 3*a - 8*b)*log(1/2*cos(
d*x + c) + 1/2) + ((3*a + 8*b)*cos(d*x + c)^4 - 2*(3*a + 8*b)*cos(d*x + c)^2 + 3*a + 8*b)*log(-1/2*cos(d*x + c
) + 1/2) - 12*b)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [B]  time = 0.24, size = 266, normalized size = 2.66 \[ \frac {4 \, {\left (3 \, a + 8 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 64 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a + b - \frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {12 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {18 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {48 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {8 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {12 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/64*(4*(3*a + 8*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 64*b*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 1)) - (a + b - 8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 12*b*(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 18*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 48*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(co
s(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 8*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)
/d

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maple [A]  time = 0.73, size = 102, normalized size = 1.02 \[ -\frac {a \cot \left (d x +c \right ) \left (\csc ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a \cot \left (d x +c \right ) \csc \left (d x +c \right )}{8 d}+\frac {3 a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {b}{4 d \sin \left (d x +c \right )^{4}}-\frac {b}{2 d \sin \left (d x +c \right )^{2}}+\frac {b \ln \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+b*sec(d*x+c)),x)

[Out]

-1/4*a*cot(d*x+c)*csc(d*x+c)^3/d-3/8*a*cot(d*x+c)*csc(d*x+c)/d+3/8/d*a*ln(csc(d*x+c)-cot(d*x+c))-1/4/d*b/sin(d
*x+c)^4-1/2/d*b/sin(d*x+c)^2+b*ln(tan(d*x+c))/d

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maxima [A]  time = 0.67, size = 110, normalized size = 1.10 \[ -\frac {{\left (3 \, a - 8 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (3 \, a \cos \left (d x + c\right )^{3} + 4 \, b \cos \left (d x + c\right )^{2} - 5 \, a \cos \left (d x + c\right ) - 6 \, b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^5*(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*((3*a - 8*b)*log(cos(d*x + c) + 1) - (3*a + 8*b)*log(cos(d*x + c) - 1) + 16*b*log(cos(d*x + c)) - 2*(3*a
*cos(d*x + c)^3 + 4*b*cos(d*x + c)^2 - 5*a*cos(d*x + c) - 6*b)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/d

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mupad [B]  time = 0.98, size = 117, normalized size = 1.17 \[ \frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a}{16}+\frac {b}{2}\right )}{d}-\frac {-\frac {3\,a\,{\cos \left (c+d\,x\right )}^3}{8}-\frac {b\,{\cos \left (c+d\,x\right )}^2}{2}+\frac {5\,a\,\cos \left (c+d\,x\right )}{8}+\frac {3\,b}{4}}{d\,\left ({\cos \left (c+d\,x\right )}^4-2\,{\cos \left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a}{16}-\frac {b}{2}\right )}{d}-\frac {b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))/sin(c + d*x)^5,x)

[Out]

(log(cos(c + d*x) - 1)*((3*a)/16 + b/2))/d - ((3*b)/4 + (5*a*cos(c + d*x))/8 - (3*a*cos(c + d*x)^3)/8 - (b*cos
(c + d*x)^2)/2)/(d*(cos(c + d*x)^4 - 2*cos(c + d*x)^2 + 1)) - (log(cos(c + d*x) + 1)*((3*a)/16 - b/2))/d - (b*
log(cos(c + d*x)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right ) \csc ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**5*(a+b*sec(d*x+c)),x)

[Out]

Integral((a + b*sec(c + d*x))*csc(c + d*x)**5, x)

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