3.27 \(\int \csc ^7(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=160 \[ -\frac {a^5}{12 d (a-a \cos (c+d x))^3}-\frac {3 a^4}{8 d (a-a \cos (c+d x))^2}-\frac {23 a^3}{16 d (a-a \cos (c+d x))}+\frac {a^3}{16 d (a \cos (c+d x)+a)}+\frac {a^2 \sec (c+d x)}{d}+\frac {9 a^2 \log (1-\cos (c+d x))}{4 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (\cos (c+d x)+1)}{4 d} \]

[Out]

-1/12*a^5/d/(a-a*cos(d*x+c))^3-3/8*a^4/d/(a-a*cos(d*x+c))^2-23/16*a^3/d/(a-a*cos(d*x+c))+1/16*a^3/d/(a+a*cos(d
*x+c))+9/4*a^2*ln(1-cos(d*x+c))/d-2*a^2*ln(cos(d*x+c))/d-1/4*a^2*ln(1+cos(d*x+c))/d+a^2*sec(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 160, 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 = {3872, 2836, 12, 88} \[ -\frac {a^5}{12 d (a-a \cos (c+d x))^3}-\frac {3 a^4}{8 d (a-a \cos (c+d x))^2}-\frac {23 a^3}{16 d (a-a \cos (c+d x))}+\frac {a^3}{16 d (a \cos (c+d x)+a)}+\frac {a^2 \sec (c+d x)}{d}+\frac {9 a^2 \log (1-\cos (c+d x))}{4 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (\cos (c+d x)+1)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^7*(a + a*Sec[c + d*x])^2,x]

[Out]

-a^5/(12*d*(a - a*Cos[c + d*x])^3) - (3*a^4)/(8*d*(a - a*Cos[c + d*x])^2) - (23*a^3)/(16*d*(a - a*Cos[c + d*x]
)) + a^3/(16*d*(a + a*Cos[c + d*x])) + (9*a^2*Log[1 - Cos[c + d*x]])/(4*d) - (2*a^2*Log[Cos[c + d*x]])/d - (a^
2*Log[1 + Cos[c + d*x]])/(4*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 88

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 2836

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*x)/b
)^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 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 ^7(c+d x) (a+a \sec (c+d x))^2 \, dx &=\int (-a-a \cos (c+d x))^2 \csc ^7(c+d x) \sec ^2(c+d x) \, dx\\ &=\frac {a^7 \operatorname {Subst}\left (\int \frac {a^2}{(-a-x)^4 x^2 (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \frac {1}{(-a-x)^4 x^2 (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \left (\frac {1}{16 a^6 (a-x)^2}+\frac {1}{4 a^7 (a-x)}+\frac {1}{a^6 x^2}-\frac {2}{a^7 x}+\frac {1}{4 a^4 (a+x)^4}+\frac {3}{4 a^5 (a+x)^3}+\frac {23}{16 a^6 (a+x)^2}+\frac {9}{4 a^7 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {a^5}{12 d (a-a \cos (c+d x))^3}-\frac {3 a^4}{8 d (a-a \cos (c+d x))^2}-\frac {23 a^3}{16 d (a-a \cos (c+d x))}+\frac {a^3}{16 d (a+a \cos (c+d x))}+\frac {9 a^2 \log (1-\cos (c+d x))}{4 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {a^2 \log (1+\cos (c+d x))}{4 d}+\frac {a^2 \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.34, size = 136, normalized size = 0.85 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (36 \csc ^4\left (\frac {1}{2} (c+d x)\right )+120 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\csc ^6\left (\frac {1}{2} (c+d x)\right ) \left (16-3 \sec ^2\left (\frac {1}{2} (c+d x)\right ) (2 \sec (c+d x)+3)\right )+48 \left (-9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 \log (\cos (c+d x))\right )\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^7*(a + a*Sec[c + d*x])^2,x]

[Out]

-1/384*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(120*Csc[(c + d*x)/2]^2 + 36*Csc[(c + d*x)/2]^4 + 48*(Log[
Cos[(c + d*x)/2]] + 4*Log[Cos[c + d*x]] - 9*Log[Sin[(c + d*x)/2]]) + Csc[(c + d*x)/2]^6*(16 - 3*Sec[(c + d*x)/
2]^2*(3 + 2*Sec[c + d*x]))))/d

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fricas [A]  time = 0.62, size = 289, normalized size = 1.81 \[ \frac {30 \, a^{2} \cos \left (d x + c\right )^{4} - 48 \, a^{2} \cos \left (d x + c\right )^{3} - 14 \, a^{2} \cos \left (d x + c\right )^{2} + 46 \, a^{2} \cos \left (d x + c\right ) - 12 \, a^{2} - 24 \, {\left (a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 27 \, {\left (a^{2} \cos \left (d x + c\right )^{5} - 2 \, a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{12 \, {\left (d \cos \left (d x + c\right )^{5} - 2 \, d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(30*a^2*cos(d*x + c)^4 - 48*a^2*cos(d*x + c)^3 - 14*a^2*cos(d*x + c)^2 + 46*a^2*cos(d*x + c) - 12*a^2 - 2
4*(a^2*cos(d*x + c)^5 - 2*a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(-cos(d*x + c)) - 3
*(a^2*cos(d*x + c)^5 - 2*a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(1/2*cos(d*x + c) +
1/2) + 27*(a^2*cos(d*x + c)^5 - 2*a^2*cos(d*x + c)^4 + 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)^5 - 2*d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2 - d*cos(d*x + c))

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giac [A]  time = 0.40, size = 238, normalized size = 1.49 \[ \frac {216 \, a^{2} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 192 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - \frac {3 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {{\left (a^{2} - \frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {90 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {396 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{3}} + \frac {192 \, {\left (2 \, a^{2} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/96*(216*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 192*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - 1)) - 3*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + (a^2 - 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c
) + 1) + 90*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 396*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)
*(cos(d*x + c) + 1)^3/(cos(d*x + c) - 1)^3 + 192*(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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maple [A]  time = 0.50, size = 121, normalized size = 0.76 \[ \frac {a^{2} \sec \left (d x +c \right )}{d}-\frac {a^{2}}{12 d \left (-1+\sec \left (d x +c \right )\right )^{3}}-\frac {5 a^{2}}{8 d \left (-1+\sec \left (d x +c \right )\right )^{2}}-\frac {39 a^{2}}{16 d \left (-1+\sec \left (d x +c \right )\right )}+\frac {9 a^{2} \ln \left (-1+\sec \left (d x +c \right )\right )}{4 d}-\frac {a^{2}}{16 d \left (1+\sec \left (d x +c \right )\right )}-\frac {a^{2} \ln \left (1+\sec \left (d x +c \right )\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^7*(a+a*sec(d*x+c))^2,x)

[Out]

a^2*sec(d*x+c)/d-1/12/d*a^2/(-1+sec(d*x+c))^3-5/8/d*a^2/(-1+sec(d*x+c))^2-39/16/d*a^2/(-1+sec(d*x+c))+9/4/d*a^
2*ln(-1+sec(d*x+c))-1/16/d*a^2/(1+sec(d*x+c))-1/4/d*a^2*ln(1+sec(d*x+c))

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maxima [A]  time = 0.33, size = 143, normalized size = 0.89 \[ -\frac {3 \, a^{2} \log \left (\cos \left (d x + c\right ) + 1\right ) - 27 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) + 24 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{4} - 24 \, a^{2} \cos \left (d x + c\right )^{3} - 7 \, a^{2} \cos \left (d x + c\right )^{2} + 23 \, a^{2} \cos \left (d x + c\right ) - 6 \, a^{2}\right )}}{\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*(3*a^2*log(cos(d*x + c) + 1) - 27*a^2*log(cos(d*x + c) - 1) + 24*a^2*log(cos(d*x + c)) - 2*(15*a^2*cos(d
*x + c)^4 - 24*a^2*cos(d*x + c)^3 - 7*a^2*cos(d*x + c)^2 + 23*a^2*cos(d*x + c) - 6*a^2)/(cos(d*x + c)^5 - 2*co
s(d*x + c)^4 + 2*cos(d*x + c)^2 - cos(d*x + c)))/d

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mupad [B]  time = 0.11, size = 147, normalized size = 0.92 \[ \frac {9\,a^2\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{4\,d}-\frac {a^2\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{4\,d}-\frac {2\,a^2\,\ln \left (\cos \left (c+d\,x\right )\right )}{d}+\frac {-\frac {5\,a^2\,{\cos \left (c+d\,x\right )}^4}{2}+4\,a^2\,{\cos \left (c+d\,x\right )}^3+\frac {7\,a^2\,{\cos \left (c+d\,x\right )}^2}{6}-\frac {23\,a^2\,\cos \left (c+d\,x\right )}{6}+a^2}{d\,\left (-{\cos \left (c+d\,x\right )}^5+2\,{\cos \left (c+d\,x\right )}^4-2\,{\cos \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/sin(c + d*x)^7,x)

[Out]

(9*a^2*log(cos(c + d*x) - 1))/(4*d) - (a^2*log(cos(c + d*x) + 1))/(4*d) - (2*a^2*log(cos(c + d*x)))/d + (a^2 -
(23*a^2*cos(c + d*x))/6 + (7*a^2*cos(c + d*x)^2)/6 + 4*a^2*cos(c + d*x)^3 - (5*a^2*cos(c + d*x)^4)/2)/(d*(cos
(c + d*x) - 2*cos(c + d*x)^2 + 2*cos(c + d*x)^4 - cos(c + d*x)^5))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**7*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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