3.25 \(\int \csc ^3(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=69 \[ -\frac {a^3}{d (a-a \cos (c+d x))}+\frac {a^2 \sec (c+d x)}{d}+\frac {2 a^2 \log (1-\cos (c+d x))}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 69, 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, 44} \[ -\frac {a^3}{d (a-a \cos (c+d x))}+\frac {a^2 \sec (c+d x)}{d}+\frac {2 a^2 \log (1-\cos (c+d x))}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

-(a^3/(d*(a - a*Cos[c + d*x]))) + (2*a^2*Log[1 - Cos[c + d*x]])/d - (2*a^2*Log[Cos[c + d*x]])/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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.56, size = 75, normalized size = 1.09 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-2 \sec (c+d x)-8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \log (\cos (c+d x))\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3*(a + a*Sec[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.57, size = 112, normalized size = 1.62 \[ \frac {2 \, a^{2} \cos \left (d x + c\right ) - a^{2} - 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

(2*a^2*cos(d*x + c) - a^2 - 2*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(-cos(d*x + c)) + 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)^2 - d*cos(d*x + c))

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giac [A]  time = 0.67, size = 135, normalized size = 1.96 \[ \frac {4 \, a^{2} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 4 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {a^{2} + \frac {5 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac {{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(4*a^2*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 4*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) - 1)) + (a^2 + 5*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + (co
s(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/d

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maple [A]  time = 0.59, size = 50, normalized size = 0.72 \[ \frac {a^{2} \sec \left (d x +c \right )}{d}-\frac {a^{2}}{d \left (-1+\sec \left (d x +c \right )\right )}+\frac {2 a^{2} \ln \left (-1+\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 68, normalized size = 0.99 \[ \frac {2 \, a^{2} \log \left (\cos \left (d x + c\right ) - 1\right ) - 2 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) + \frac {2 \, a^{2} \cos \left (d x + c\right ) - a^{2}}{\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.90, size = 61, normalized size = 0.88 \[ -\frac {2\,a^2\,\cos \left (c+d\,x\right )-a^2}{d\,\left (\cos \left (c+d\,x\right )-{\cos \left (c+d\,x\right )}^2\right )}-\frac {4\,a^2\,\mathrm {atanh}\left (2\,\cos \left (c+d\,x\right )-1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/sin(c + d*x)^3,x)

[Out]

- (2*a^2*cos(c + d*x) - a^2)/(d*(cos(c + d*x) - cos(c + d*x)^2)) - (4*a^2*atanh(2*cos(c + d*x) - 1))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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