3.46 \(\int \csc ^7(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=157 \[ -\frac {a^6}{6 d (a-a \cos (c+d x))^3}-\frac {7 a^5}{8 d (a-a \cos (c+d x))^2}-\frac {31 a^4}{8 d (a-a \cos (c+d x))}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {111 a^3 \log (1-\cos (c+d x))}{16 d}-\frac {7 a^3 \log (\cos (c+d x))}{d}+\frac {a^3 \log (\cos (c+d x)+1)}{16 d} \]

[Out]

-1/6*a^6/d/(a-a*cos(d*x+c))^3-7/8*a^5/d/(a-a*cos(d*x+c))^2-31/8*a^4/d/(a-a*cos(d*x+c))+111/16*a^3*ln(1-cos(d*x
+c))/d-7*a^3*ln(cos(d*x+c))/d+1/16*a^3*ln(1+cos(d*x+c))/d+3*a^3*sec(d*x+c)/d+1/2*a^3*sec(d*x+c)^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 157, 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} \[ -\frac {a^6}{6 d (a-a \cos (c+d x))^3}-\frac {7 a^5}{8 d (a-a \cos (c+d x))^2}-\frac {31 a^4}{8 d (a-a \cos (c+d x))}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {111 a^3 \log (1-\cos (c+d x))}{16 d}-\frac {7 a^3 \log (\cos (c+d x))}{d}+\frac {a^3 \log (\cos (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^7*(a + a*Sec[c + d*x])^3,x]

[Out]

-a^6/(6*d*(a - a*Cos[c + d*x])^3) - (7*a^5)/(8*d*(a - a*Cos[c + d*x])^2) - (31*a^4)/(8*d*(a - a*Cos[c + d*x]))
+ (111*a^3*Log[1 - Cos[c + d*x]])/(16*d) - (7*a^3*Log[Cos[c + d*x]])/d + (a^3*Log[1 + Cos[c + d*x]])/(16*d) +
(3*a^3*Sec[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(2*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))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^7(c+d x) \sec ^3(c+d x) \, dx\\ &=\frac {a^7 \operatorname {Subst}\left (\int \frac {a^3}{(-a-x)^4 x^3 (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^{10} \operatorname {Subst}\left (\int \frac {1}{(-a-x)^4 x^3 (-a+x)} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^{10} \operatorname {Subst}\left (\int \left (-\frac {1}{16 a^7 (a-x)}-\frac {1}{a^5 x^3}+\frac {3}{a^6 x^2}-\frac {7}{a^7 x}+\frac {1}{2 a^4 (a+x)^4}+\frac {7}{4 a^5 (a+x)^3}+\frac {31}{8 a^6 (a+x)^2}+\frac {111}{16 a^7 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {a^6}{6 d (a-a \cos (c+d x))^3}-\frac {7 a^5}{8 d (a-a \cos (c+d x))^2}-\frac {31 a^4}{8 d (a-a \cos (c+d x))}+\frac {111 a^3 \log (1-\cos (c+d x))}{16 d}-\frac {7 a^3 \log (\cos (c+d x))}{d}+\frac {a^3 \log (1+\cos (c+d x))}{16 d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.06, size = 129, normalized size = 0.82 \[ -\frac {a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (2 \csc ^6\left (\frac {1}{2} (c+d x)\right )+21 \csc ^4\left (\frac {1}{2} (c+d x)\right )+186 \csc ^2\left (\frac {1}{2} (c+d x)\right )-12 \left (4 \sec ^2(c+d x)+24 \sec (c+d x)+111 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-56 \log (\cos (c+d x))\right )\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^7*(a + a*Sec[c + d*x])^3,x]

[Out]

-1/768*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(186*Csc[(c + d*x)/2]^2 + 21*Csc[(c + d*x)/2]^4 + 2*Csc[(c
+ d*x)/2]^6 - 12*(Log[Cos[(c + d*x)/2]] - 56*Log[Cos[c + d*x]] + 111*Log[Sin[(c + d*x)/2]] + 24*Sec[c + d*x]
+ 4*Sec[c + d*x]^2)))/d

________________________________________________________________________________________

fricas [B]  time = 0.67, size = 297, normalized size = 1.89 \[ \frac {330 \, a^{3} \cos \left (d x + c\right )^{4} - 822 \, a^{3} \cos \left (d x + c\right )^{3} + 596 \, a^{3} \cos \left (d x + c\right )^{2} - 72 \, a^{3} \cos \left (d x + c\right ) - 24 \, a^{3} - 336 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + 3 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 333 \, {\left (a^{3} \cos \left (d x + c\right )^{5} - 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{5} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/48*(330*a^3*cos(d*x + c)^4 - 822*a^3*cos(d*x + c)^3 + 596*a^3*cos(d*x + c)^2 - 72*a^3*cos(d*x + c) - 24*a^3
- 336*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-cos(d*x + c
)) + 3*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(1/2*cos(d*x
+ c) + 1/2) + 333*(a^3*cos(d*x + c)^5 - 3*a^3*cos(d*x + c)^4 + 3*a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log
(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^5 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^3 - d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [A]  time = 0.76, size = 243, normalized size = 1.55 \[ \frac {666 \, a^{3} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 672 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {{\left (2 \, a^{3} - \frac {27 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {234 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1221 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{3}} + \frac {48 \, {\left (33 \, a^{3} + \frac {50 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {21 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(666*a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 672*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - 1)) + (2*a^3 - 27*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 234*a^3*(cos(d*x + c) - 1)^2/(cos(d
*x + c) + 1)^2 - 1221*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)*(cos(d*x + c) + 1)^3/(cos(d*x + c) - 1)^3
+ 48*(33*a^3 + 50*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 21*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^
2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.61, size = 120, normalized size = 0.76 \[ \frac {a^{3} \left (\sec ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \sec \left (d x +c \right )}{d}-\frac {a^{3}}{6 d \left (-1+\sec \left (d x +c \right )\right )^{3}}-\frac {11 a^{3}}{8 d \left (-1+\sec \left (d x +c \right )\right )^{2}}-\frac {49 a^{3}}{8 d \left (-1+\sec \left (d x +c \right )\right )}+\frac {111 a^{3} \ln \left (-1+\sec \left (d x +c \right )\right )}{16 d}+\frac {a^{3} \ln \left (1+\sec \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^7*(a+a*sec(d*x+c))^3,x)

[Out]

1/2*a^3*sec(d*x+c)^2/d+3*a^3*sec(d*x+c)/d-1/6/d*a^3/(-1+sec(d*x+c))^3-11/8/d*a^3/(-1+sec(d*x+c))^2-49/8/d*a^3/
(-1+sec(d*x+c))+111/16/d*a^3*ln(-1+sec(d*x+c))+1/16/d*a^3*ln(1+sec(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 145, normalized size = 0.92 \[ \frac {3 \, a^{3} \log \left (\cos \left (d x + c\right ) + 1\right ) + 333 \, a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - 336 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, {\left (165 \, a^{3} \cos \left (d x + c\right )^{4} - 411 \, a^{3} \cos \left (d x + c\right )^{3} + 298 \, a^{3} \cos \left (d x + c\right )^{2} - 36 \, a^{3} \cos \left (d x + c\right ) - 12 \, a^{3}\right )}}{\cos \left (d x + c\right )^{5} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^7*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/48*(3*a^3*log(cos(d*x + c) + 1) + 333*a^3*log(cos(d*x + c) - 1) - 336*a^3*log(cos(d*x + c)) + 2*(165*a^3*cos
(d*x + c)^4 - 411*a^3*cos(d*x + c)^3 + 298*a^3*cos(d*x + c)^2 - 36*a^3*cos(d*x + c) - 12*a^3)/(cos(d*x + c)^5
- 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 - cos(d*x + c)^2))/d

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 151, normalized size = 0.96 \[ \frac {111\,a^3\,\ln \left (\cos \left (c+d\,x\right )-1\right )}{16\,d}+\frac {a^3\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{16\,d}+\frac {-\frac {55\,a^3\,{\cos \left (c+d\,x\right )}^4}{8}+\frac {137\,a^3\,{\cos \left (c+d\,x\right )}^3}{8}-\frac {149\,a^3\,{\cos \left (c+d\,x\right )}^2}{12}+\frac {3\,a^3\,\cos \left (c+d\,x\right )}{2}+\frac {a^3}{2}}{d\,\left (-{\cos \left (c+d\,x\right )}^5+3\,{\cos \left (c+d\,x\right )}^4-3\,{\cos \left (c+d\,x\right )}^3+{\cos \left (c+d\,x\right )}^2\right )}-\frac {7\,a^3\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^3/sin(c + d*x)^7,x)

[Out]

(111*a^3*log(cos(c + d*x) - 1))/(16*d) + (a^3*log(cos(c + d*x) + 1))/(16*d) + ((3*a^3*cos(c + d*x))/2 + a^3/2
- (149*a^3*cos(c + d*x)^2)/12 + (137*a^3*cos(c + d*x)^3)/8 - (55*a^3*cos(c + d*x)^4)/8)/(d*(cos(c + d*x)^2 - 3
*cos(c + d*x)^3 + 3*cos(c + d*x)^4 - cos(c + d*x)^5)) - (7*a^3*log(cos(c + d*x)))/d

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________