3.1.8 \(\int \csc ^5(c+d x) (a+a \sec (c+d x)) \, dx\) [8]

Optimal. Leaf size=118 \[ -\frac {a^3}{8 d (a-a \cos (c+d x))^2}-\frac {a^2}{2 d (a-a \cos (c+d x))}-\frac {a^2}{8 d (a+a \cos (c+d x))}+\frac {11 a \log (1-\cos (c+d x))}{16 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {5 a \log (1+\cos (c+d x))}{16 d} \]

[Out]

-1/8*a^3/d/(a-a*cos(d*x+c))^2-1/2*a^2/d/(a-a*cos(d*x+c))-1/8*a^2/d/(a+a*cos(d*x+c))+11/16*a*ln(1-cos(d*x+c))/d
-a*ln(cos(d*x+c))/d+5/16*a*ln(1+cos(d*x+c))/d

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Rubi [A]
time = 0.09, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3957, 2915, 12, 90} \begin {gather*} -\frac {a^3}{8 d (a-a \cos (c+d x))^2}-\frac {a^2}{2 d (a-a \cos (c+d x))}-\frac {a^2}{8 d (a \cos (c+d x)+a)}+\frac {11 a \log (1-\cos (c+d x))}{16 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {5 a \log (\cos (c+d x)+1)}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*a^3/(d*(a - a*Cos[c + d*x])^2) - a^2/(2*d*(a - a*Cos[c + d*x])) - a^2/(8*d*(a + a*Cos[c + d*x])) + (11*a*
Log[1 - Cos[c + d*x]])/(16*d) - (a*Log[Cos[c + d*x]])/d + (5*a*Log[1 + Cos[c + d*x]])/(16*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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+a \sec (c+d x)) \, dx &=-\int (-a-a \cos (c+d x)) \csc ^5(c+d x) \sec (c+d x) \, dx\\ &=\frac {a^5 \text {Subst}\left (\int \frac {a}{(-a-x)^3 x (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^6 \text {Subst}\left (\int \frac {1}{(-a-x)^3 x (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^6 \text {Subst}\left (\int \left (-\frac {1}{8 a^4 (a-x)^2}-\frac {5}{16 a^5 (a-x)}-\frac {1}{a^5 x}+\frac {1}{4 a^3 (a+x)^3}+\frac {1}{2 a^4 (a+x)^2}+\frac {11}{16 a^5 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {a^3}{8 d (a-a \cos (c+d x))^2}-\frac {a^2}{2 d (a-a \cos (c+d x))}-\frac {a^2}{8 d (a+a \cos (c+d x))}+\frac {11 a \log (1-\cos (c+d x))}{16 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {5 a \log (1+\cos (c+d x))}{16 d}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 164, normalized size = 1.39 \begin {gather*} -\frac {3 a \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {a \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {3 a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {a \left (2 \csc ^2(c+d x)+\csc ^4(c+d x)+4 \log (\cos (c+d x))-4 \log (\sin (c+d x))\right )}{4 d}+\frac {3 a \sec ^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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5*(a + a*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) - (a*(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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Maple [A]
time = 0.12, size = 66, normalized size = 0.56

method	result	size
derivativedivides	\(-\frac {a \left (\frac {1}{8 \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {3}{4 \left (-1+\sec \left (d x +c \right )\right )}-\frac {11 \ln \left (-1+\sec \left (d x +c \right )\right )}{16}-\frac {1}{8 \left (1+\sec \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sec \left (d x +c \right )\right )}{16}\right )}{d}\)	\(66\)
default	\(-\frac {a \left (\frac {1}{8 \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {3}{4 \left (-1+\sec \left (d x +c \right )\right )}-\frac {11 \ln \left (-1+\sec \left (d x +c \right )\right )}{16}-\frac {1}{8 \left (1+\sec \left (d x +c \right )\right )}-\frac {5 \ln \left (1+\sec \left (d x +c \right )\right )}{16}\right )}{d}\)	\(66\)
norman	\(\frac {-\frac {a}{32 d}-\frac {5 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {11 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(107\)
risch	\(\frac {a \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}-18 \,{\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2}}+\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d}+\frac {11 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d*a*(1/8/(-1+sec(d*x+c))^2+3/4/(-1+sec(d*x+c))-11/16*ln(-1+sec(d*x+c))-1/8/(1+sec(d*x+c))-5/16*ln(1+sec(d*x
+c)))

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Maxima [A]
time = 0.26, size = 95, normalized size = 0.81 \begin {gather*} \frac {5 \, a \log \left (\cos \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\cos \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - 6 \, a\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 1}}{16 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(5*a*log(cos(d*x + c) + 1) + 11*a*log(cos(d*x + c) - 1) - 16*a*log(cos(d*x + c)) + 2*(3*a*cos(d*x + c)^2
+ a*cos(d*x + c) - 6*a)/(cos(d*x + c)^3 - cos(d*x + c)^2 - cos(d*x + c) + 1))/d

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Fricas [A]
time = 3.14, size = 193, normalized size = 1.64 \begin {gather*} \frac {6 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - 16 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\cos \left (d x + c\right )\right ) + 5 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 11 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - 16*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*lo
g(-cos(d*x + c)) + 5*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(1/2*cos(d*x + c) + 1/2) +
11*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(-1/2*cos(d*x + c) + 1/2) - 12*a)/(d*cos(d*x
+ c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 149, normalized size = 1.26 \begin {gather*} \frac {22 \, a \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 32 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a - \frac {10 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {33 \, a {\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 {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{32 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(22*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 32*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1)) - (a - 10*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 33*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)
*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 + 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/d

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Mupad [B]
time = 0.10, size = 99, normalized size = 0.84 \begin {gather*} \frac {11\,a\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{16\,d}-\frac {a\,\ln \left (\cos \left (c+d\,x\right )\right )}{d}+\frac {5\,a\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{16\,d}+\frac {\frac {3\,a\,{\cos \left (c+d\,x\right )}^2}{8}+\frac {a\,\cos \left (c+d\,x\right )}{8}-\frac {3\,a}{4}}{d\,\left ({\cos \left (c+d\,x\right )}^3-\cos \left (c+d\,x\right )+{\sin \left (c+d\,x\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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