3.34 \(\int \csc ^4(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=87 \[ -\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}+\frac {10 a^2 \tan (c+d x)}{3 d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))} \]

[Out]

2*a^2*arctanh(sin(d*x+c))/d+10/3*a^2*tan(d*x+c)/d-2*a^2*tan(d*x+c)/d/(1-cos(d*x+c))-1/3*a^4*tan(d*x+c)/d/(a-a*
cos(d*x+c))^2

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Rubi [A]  time = 0.30, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2869, 2766, 2978, 2748, 3767, 8, 3770} \[ \frac {10 a^2 \tan (c+d x)}{3 d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

(2*a^2*ArcTanh[Sin[c + d*x]])/d + (10*a^2*Tan[c + d*x])/(3*d) - (2*a^2*Tan[c + d*x])/(d*(1 - Cos[c + d*x])) -
(a^4*Tan[c + d*x])/(3*d*(a - a*Cos[c + d*x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 ^4(c+d x) (a+a \sec (c+d x))^2 \, dx &=\int (-a-a \cos (c+d x))^2 \csc ^4(c+d x) \sec ^2(c+d x) \, dx\\ &=a^4 \int \frac {\sec ^2(c+d x)}{(-a+a \cos (c+d x))^2} \, dx\\ &=-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}+\frac {1}{3} a^2 \int \frac {(-4 a-2 a \cos (c+d x)) \sec ^2(c+d x)}{-a+a \cos (c+d x)} \, dx\\ &=-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}+\frac {1}{3} \int \left (10 a^2+6 a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}+\left (2 a^2\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (10 a^2\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}-\frac {\left (10 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {10 a^2 \tan (c+d x)}{3 d}-\frac {2 a^2 \tan (c+d x)}{d (1-\cos (c+d x))}-\frac {a^4 \tan (c+d x)}{3 d (a-a \cos (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 1.85, size = 228, normalized size = 2.62 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\cot \left (\frac {c}{2}\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )+6 \left (\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\left (\csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) (7 \cos (c+d x)-8) \csc ^3\left (\frac {1}{2} (c+d x)\right )\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-(Cot[c/2]*Csc[(c + d*x)/2]^2) - (-8 + 7*Cos[c + d*x])*Csc[c/2]*
Csc[(c + d*x)/2]^3*Sin[(d*x)/2] + 6*(-2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))))/(24*d)

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fricas [A]  time = 0.69, size = 159, normalized size = 1.83 \[ -\frac {10 \, a^{2} \cos \left (d x + c\right )^{3} - 4 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, a^{2}}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(10*a^2*cos(d*x + c)^3 - 4*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c) - 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x
+ c))*log(sin(d*x + c) + 1)*sin(d*x + c) + 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c))*log(-sin(d*x + c) + 1)*s
in(d*x + c) + 3*a^2)/((d*cos(d*x + c)^2 - d*cos(d*x + c))*sin(d*x + c))

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giac [A]  time = 0.30, size = 104, normalized size = 1.20 \[ \frac {12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(12*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 12*a^2*tan(1/2*d*
x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (15*a^2*tan(1/2*d*x + 1/2*c)^2 + a^2)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.88, size = 140, normalized size = 1.61 \[ -\frac {10 a^{2} \cot \left (d x +c \right )}{3 d}-\frac {a^{2} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{2}}{3 d \sin \left (d x +c \right )^{3}}-\frac {2 a^{2}}{d \sin \left (d x +c \right )}+\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {a^{2}}{3 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4 a^{2}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*(a+a*sec(d*x+c))^2,x)

[Out]

-10/3*a^2*cot(d*x+c)/d-1/3/d*a^2*cot(d*x+c)*csc(d*x+c)^2-2/3/d*a^2/sin(d*x+c)^3-2/d*a^2/sin(d*x+c)+2/d*a^2*ln(
sec(d*x+c)+tan(d*x+c))-1/3/d*a^2/sin(d*x+c)^3/cos(d*x+c)+4/3/d*a^2/sin(d*x+c)/cos(d*x+c)

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maxima [A]  time = 0.34, size = 113, normalized size = 1.30 \[ -\frac {a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + a^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(a^2*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + a^2*
((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + (3*tan(d*x + c)^2 + 1)*a^2/tan(d*x + c)^3)/d

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mupad [B]  time = 2.48, size = 91, normalized size = 1.05 \[ \frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {a^2}{3}}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/sin(c + d*x)^4,x)

[Out]

(4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - ((14*a^2*tan(c/2 + (d*x)/2)^2)/3 - 9*a^2*tan(c/2 + (d*x)/2)^4 + a^2/3)/(
d*(2*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2 + (d*x)/2)^5))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \csc ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*csc(c + d*x)**4*sec(c + d*x), x) + Integral(csc(c + d*x)**4*sec(c + d*x)**2, x) + Integral(cs
c(c + d*x)**4, x))

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