3.1.44 \(\int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\) [44]

Optimal. Leaf size=88 \[ -\frac {2 a^4}{d (a-a \cos (c+d x))}+\frac {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 88, 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, 78} \begin {gather*} -\frac {2 a^4}{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 {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^4)/(d*(a - a*Cos[c + d*x])) + (5*a^3*Log[1 - Cos[c + d*x]])/d - (5*a^3*Log[Cos[c + d*x]])/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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 ^3(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^3(c+d x) \sec ^3(c+d x) \, dx\\ &=\frac {a^3 \text {Subst}\left (\int \frac {a^3 (-a+x)}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^6 \text {Subst}\left (\int \frac {-a+x}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac {a^6 \text {Subst}\left (\int \left (-\frac {1}{a x^3}+\frac {3}{a^2 x^2}-\frac {5}{a^3 x}+\frac {2}{a^2 (a+x)^2}+\frac {5}{a^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac {2 a^4}{d (a-a \cos (c+d x))}+\frac {5 a^3 \log (1-\cos (c+d x))}{d}-\frac {5 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 88, normalized size = 1.00 \begin {gather*} -\frac {a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+10 \left (\log (\cos (c+d x))-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-6 \sec (c+d x)-\sec ^2(c+d x)\right )}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*Csc[(c + d*x)/2]^2 + 10*(Log[Cos[c + d*x]] - 2*Log[Sin[(
c + d*x)/2]]) - 6*Sec[c + d*x] - Sec[c + d*x]^2))/d

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Maple [A]
time = 0.12, size = 50, normalized size = 0.57

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{2}+3 \sec \left (d x +c \right )-\frac {2}{-1+\sec \left (d x +c \right )}+5 \ln \left (-1+\sec \left (d x +c \right )\right )\right )}{d}\)	\(50\)
default	\(\frac {a^{3} \left (\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{2}+3 \sec \left (d x +c \right )-\frac {2}{-1+\sec \left (d x +c \right )}+5 \ln \left (-1+\sec \left (d x +c \right )\right )\right )}{d}\)	\(50\)
risch	\(\frac {2 a^{3} \left (5 \,{\mathrm e}^{5 i \left (d x +c \right )}-5 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{3 i \left (d x +c \right )}-5 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}+\frac {10 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(132\)
norman	\(\frac {\frac {8 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3}}{d}-\frac {5 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {10 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 84, normalized size = 0.95 \begin {gather*} \frac {10 \, a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - 10 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac {10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*log(cos(d*x + c) - 1) - 10*a^3*log(cos(d*x + c)) + (10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a
^3)/(cos(d*x + c)^3 - cos(d*x + c)^2))/d

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Fricas [A]
time = 3.54, size = 132, normalized size = 1.50 \begin {gather*} \frac {10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3} - 10 \, {\left (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 ) + 10 \, {\left (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 )}{2 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a^3 - 10*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-cos(
d*x + c)) + 10*(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)^3 - d*c
os(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \csc ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \csc ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \csc ^{3}{\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)**3*(a+a*sec(d*x+c))**3,x)

[Out]

a**3*(Integral(3*csc(c + d*x)**3*sec(c + d*x), x) + Integral(3*csc(c + d*x)**3*sec(c + d*x)**2, x) + Integral(
csc(c + d*x)**3*sec(c + d*x)**3, x) + Integral(csc(c + d*x)**3, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (87) = 174\).
time = 0.54, size = 189, normalized size = 2.15 \begin {gather*} \frac {10 \, a^{3} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 10 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {2 \, {\left (a^{3} - \frac {5 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} + \frac {27 \, a^{3} + \frac {38 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {15 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 10*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + 2*(a^3 - 5*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) +
(27*a^3 + 38*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 15*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d

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Mupad [B]
time = 0.10, size = 75, normalized size = 0.85 \begin {gather*} \frac {-5\,a^3\,{\cos \left (c+d\,x\right )}^2+\frac {5\,a^3\,\cos \left (c+d\,x\right )}{2}+\frac {a^3}{2}}{d\,\left ({\cos \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^3\right )}-\frac {10\,a^3\,\mathrm {atanh}\left (2\,\cos \left (c+d\,x\right )-1\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*a^3*cos(c + d*x))/2 + a^3/2 - 5*a^3*cos(c + d*x)^2)/(d*(cos(c + d*x)^2 - cos(c + d*x)^3)) - (10*a^3*atanh(
2*cos(c + d*x) - 1))/d

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