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

Optimal. Leaf size=80 \[ -\frac {3 a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {9 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

-9/2*a^3*arctanh(cos(d*x+c))/d-3*a^3*cot(d*x+c)/d-1/2*a^3*cot(d*x+c)*csc(d*x+c)/d+4*a^3*cos(d*x+c)/d/(1-sin(d*
x+c))

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Rubi [A]  time = 0.16, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2872, 3770, 3767, 8, 3768, 2648} \[ -\frac {3 a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {9 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (3*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d) + (4*a^3
*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))

Rule 8

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin {align*} \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=a^2 \int \left (4 a \csc (c+d x)+3 a \csc ^2(c+d x)+a \csc ^3(c+d x)-\frac {4 a}{-1+\sin (c+d x)}\right ) \, dx\\ &=a^3 \int \csc ^3(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (4 a^3\right ) \int \csc (c+d x) \, dx-\left (4 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=-\frac {4 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {1}{2} a^3 \int \csc (c+d x) \, dx-\frac {\left (3 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac {9 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 124, normalized size = 1.55 \[ \frac {a^3 \left (12 \tan \left (\frac {1}{2} (c+d x)\right )-12 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+36 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-36 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {64 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-12*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 36*Log[Cos[(c + d*x)/2]] + 36*Log[Sin[(c + d*x)/2]] + Sec[(c
+ d*x)/2]^2 + (64*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 12*Tan[(c + d*x)/2]))/(8*d)

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fricas [B]  time = 0.47, size = 300, normalized size = 3.75 \[ \frac {28 \, a^{3} \cos \left (d x + c\right )^{3} + 18 \, a^{3} \cos \left (d x + c\right )^{2} - 26 \, a^{3} \cos \left (d x + c\right ) - 16 \, a^{3} - 9 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 5 \, a^{3} \cos \left (d x + c\right ) - 8 \, a^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(28*a^3*cos(d*x + c)^3 + 18*a^3*cos(d*x + c)^2 - 26*a^3*cos(d*x + c) - 16*a^3 - 9*(a^3*cos(d*x + c)^3 + a^
3*cos(d*x + c)^2 - a^3*cos(d*x + c) - a^3 - (a^3*cos(d*x + c)^2 - a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/
2) + 9*(a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - a^3 - (a^3*cos(d*x + c)^2 - a^3)*sin(d*x
+ c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(14*a^3*cos(d*x + c)^2 + 5*a^3*cos(d*x + c) - 8*a^3)*sin(d*x + c))/(d*c
os(d*x + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) - (d*cos(d*x + c)^2 - d)*sin(d*x + c) - d)

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giac [A]  time = 0.23, size = 116, normalized size = 1.45 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {64 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} - \frac {54 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^3*tan(1/2*d*x + 1/2*c) - 64*a^3
/(tan(1/2*d*x + 1/2*c) - 1) - (54*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x
+ 1/2*c)^2)/d

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maple [A]  time = 0.70, size = 117, normalized size = 1.46 \[ \frac {a^{3} \tan \left (d x +c \right )}{d}+\frac {9 a^{3}}{2 d \cos \left (d x +c \right )}+\frac {9 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {6 a^{3} \cot \left (d x +c \right )}{d}-\frac {a^{3}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*tan(d*x+c)/d+9/2/d*a^3/cos(d*x+c)+9/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))+3/d*a^3/sin(d*x+c)/cos(d*x+c)-6*a^3*
cot(d*x+c)/d-1/2/d*a^3/sin(d*x+c)^2/cos(d*x+c)

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maxima [A]  time = 0.34, size = 135, normalized size = 1.69 \[ \frac {a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + 4 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^3*(2*(3*cos(d*x + c)^2 - 2)/(cos(d*x + c)^3 - cos(d*x + c)) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x +
c) - 1)) + 6*a^3*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a^3*(1/tan(d*x + c) -
tan(d*x + c)) + 4*a^3*tan(d*x + c))/d

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mupad [B]  time = 8.95, size = 125, normalized size = 1.56 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {9\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {-38\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {11\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (9*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) + (3*a^3*tan(c/2 + (d*x)/2))/(2*d) -
(a^3/2 - 38*a^3*tan(c/2 + (d*x)/2)^2 + (11*a^3*tan(c/2 + (d*x)/2))/2)/(d*(4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 +
(d*x)/2)^3))

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

[Out]

Timed out

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