3.1.26 \(\int \csc ^5(c+d x) (a+a \sec (c+d x))^2 \, dx\) [26]

Optimal. Leaf size=115 \[ -\frac {a^4}{4 d (a-a \cos (c+d x))^2}-\frac {5 a^3}{4 d (a-a \cos (c+d x))}+\frac {17 a^2 \log (1-\cos (c+d x))}{8 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (1+\cos (c+d x))}{8 d}+\frac {a^2 \sec (c+d x)}{d} \]

[Out]

-1/4*a^4/d/(a-a*cos(d*x+c))^2-5/4*a^3/d/(a-a*cos(d*x+c))+17/8*a^2*ln(1-cos(d*x+c))/d-2*a^2*ln(cos(d*x+c))/d-1/
8*a^2*ln(1+cos(d*x+c))/d+a^2*sec(d*x+c)/d

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Rubi [A]
time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 90} \begin {gather*} -\frac {a^4}{4 d (a-a \cos (c+d x))^2}-\frac {5 a^3}{4 d (a-a \cos (c+d x))}+\frac {a^2 \sec (c+d x)}{d}+\frac {17 a^2 \log (1-\cos (c+d x))}{8 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (\cos (c+d x)+1)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^5*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/4*a^4/(d*(a - a*Cos[c + d*x])^2) - (5*a^3)/(4*d*(a - a*Cos[c + d*x])) + (17*a^2*Log[1 - Cos[c + d*x]])/(8*d
) - (2*a^2*Log[Cos[c + d*x]])/d - (a^2*Log[1 + Cos[c + d*x]])/(8*d) + (a^2*Sec[c + d*x])/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))^2 \, dx &=\int (-a-a \cos (c+d x))^2 \csc ^5(c+d x) \sec ^2(c+d x) \, dx\\ &=\frac {a^5 \text {Subst}\left (\int \frac {a^2}{(-a-x)^3 x^2 (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^7 \text {Subst}\left (\int \frac {1}{(-a-x)^3 x^2 (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^7 \text {Subst}\left (\int \left (\frac {1}{8 a^5 (a-x)}+\frac {1}{a^4 x^2}-\frac {2}{a^5 x}+\frac {1}{2 a^3 (a+x)^3}+\frac {5}{4 a^4 (a+x)^2}+\frac {17}{8 a^5 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {a^4}{4 d (a-a \cos (c+d x))^2}-\frac {5 a^3}{4 d (a-a \cos (c+d x))}+\frac {17 a^2 \log (1-\cos (c+d x))}{8 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (1+\cos (c+d x))}{8 d}+\frac {a^2 \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 103, normalized size = 0.90 \begin {gather*} -\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (10 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\csc ^4\left (\frac {1}{2} (c+d x)\right )+4 \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 \log (\cos (c+d x))-17 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 \sec (c+d x)\right )\right )}{64 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^5*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/64*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(10*Csc[(c + d*x)/2]^2 + Csc[(c + d*x)/2]^4 + 4*(Log[Cos[(c
+ d*x)/2]] + 8*Log[Cos[c + d*x]] - 17*Log[Sin[(c + d*x)/2]] - 4*Sec[c + d*x])))/d

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Maple [A]
time = 0.12, size = 64, normalized size = 0.56

method	result	size
derivativedivides	\(-\frac {a^{2} \left (-\sec \left (d x +c \right )+\frac {1}{4 \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {7}{4 \left (-1+\sec \left (d x +c \right )\right )}-\frac {17 \ln \left (-1+\sec \left (d x +c \right )\right )}{8}+\frac {\ln \left (1+\sec \left (d x +c \right )\right )}{8}\right )}{d}\)	\(64\)
default	\(-\frac {a^{2} \left (-\sec \left (d x +c \right )+\frac {1}{4 \left (-1+\sec \left (d x +c \right )\right )^{2}}+\frac {7}{4 \left (-1+\sec \left (d x +c \right )\right )}-\frac {17 \ln \left (-1+\sec \left (d x +c \right )\right )}{8}+\frac {\ln \left (1+\sec \left (d x +c \right )\right )}{8}\right )}{d}\)	\(64\)
norman	\(\frac {\frac {a^{2}}{16 d}+\frac {11 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d}-\frac {11 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {17 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(134\)
risch	\(\frac {a^{2} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}-28 \,{\mathrm e}^{4 i \left (d x +c \right )}+34 \,{\mathrm e}^{3 i \left (d x +c \right )}-28 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {17 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{4 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^5*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/d*a^2*(-sec(d*x+c)+1/4/(-1+sec(d*x+c))^2+7/4/(-1+sec(d*x+c))-17/8*ln(-1+sec(d*x+c))+1/8*ln(1+sec(d*x+c)))

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Maxima [A]
time = 0.29, size = 104, normalized size = 0.90 \begin {gather*} -\frac {a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) - 17 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) + 16 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (9 \, a^{2} \cos \left (d x + c\right )^{2} - 14 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )}}{\cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )}}{8 \, 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))^2,x, algorithm="maxima")

[Out]

-1/8*(a^2*log(cos(d*x + c) + 1) - 17*a^2*log(cos(d*x + c) - 1) + 16*a^2*log(cos(d*x + c)) - 2*(9*a^2*cos(d*x +
c)^2 - 14*a^2*cos(d*x + c) + 4*a^2)/(cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c)))/d

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Fricas [A]
time = 2.79, size = 209, normalized size = 1.82 \begin {gather*} \frac {18 \, a^{2} \cos \left (d x + c\right )^{2} - 28 \, a^{2} \cos \left (d x + c\right ) + 8 \, a^{2} - 16 \, {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 17 \, {\left (a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\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))^2,x, algorithm="fricas")

[Out]

1/8*(18*a^2*cos(d*x + c)^2 - 28*a^2*cos(d*x + c) + 8*a^2 - 16*(a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2
*cos(d*x + c))*log(-cos(d*x + c)) - (a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(1/2*cos
(d*x + c) + 1/2) + 17*(a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1
/2))/(d*cos(d*x + c)^3 - 2*d*cos(d*x + c)^2 + d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \csc ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{5}{\left (c + d x \right )} \sec ^{2}{\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))**2,x)

[Out]

a**2*(Integral(2*csc(c + d*x)**5*sec(c + d*x), x) + Integral(csc(c + d*x)**5*sec(c + d*x)**2, x) + Integral(cs
c(c + d*x)**5, x))

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Giac [A]
time = 0.54, size = 191, normalized size = 1.66 \begin {gather*} \frac {34 \, a^{2} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 32 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {{\left (a^{2} - \frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {51 \, a^{2} {\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 {32 \, {\left (2 \, a^{2} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 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))^2,x, algorithm="giac")

[Out]

1/16*(34*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 32*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) - (a^2 - 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 51*a^2*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 + 32*(2*a^2 + a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/
((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/d

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Mupad [B]
time = 0.10, size = 109, normalized size = 0.95 \begin {gather*} \frac {17\,a^2\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{8\,d}-\frac {a^2\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{8\,d}+\frac {\frac {9\,a^2\,{\cos \left (c+d\,x\right )}^2}{4}-\frac {7\,a^2\,\cos \left (c+d\,x\right )}{2}+a^2}{d\,\left ({\cos \left (c+d\,x\right )}^3-2\,{\cos \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\right )}-\frac {2\,a^2\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/sin(c + d*x)^5,x)

[Out]

(17*a^2*log(cos(c + d*x) - 1))/(8*d) - (a^2*log(cos(c + d*x) + 1))/(8*d) + (a^2 - (7*a^2*cos(c + d*x))/2 + (9*
a^2*cos(c + d*x)^2)/4)/(d*(cos(c + d*x) - 2*cos(c + d*x)^2 + cos(c + d*x)^3)) - (2*a^2*log(cos(c + d*x)))/d

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