3.771 \(\int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=98 \[ -\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {5 a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

-11/2*a^3*arctanh(cos(d*x+c))/d-5*a^3*cot(d*x+c)/d-1/3*a^3*cot(d*x+c)^3/d-3/2*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 = 0.18, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {5 a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(-11*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/(3*d) - (3*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 ^4(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)+4 a \csc ^2(c+d x)+3 a \csc ^3(c+d x)+a \csc ^4(c+d x)-\frac {4 a}{-1+\sin (c+d x)}\right ) \, dx\\ &=a^3 \int \csc ^4(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^3(c+d x) \, dx+\left (4 a^3\right ) \int \csc (c+d x) \, dx+\left (4 a^3\right ) \int \csc ^2(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 {3 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} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac {\left (4 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac {11 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {5 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 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 [B]  time = 6.14, size = 211, normalized size = 2.15 \[ a^3 \left (\frac {7 \tan \left (\frac {1}{2} (c+d x)\right )}{3 d}-\frac {7 \cot \left (\frac {1}{2} (c+d x)\right )}{3 d}-\frac {3 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {3 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {11 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {11 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {8 \sin \left (\frac {1}{2} (c+d x)\right )}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

a^3*((-7*Cot[(c + d*x)/2])/(3*d) - (3*Csc[(c + d*x)/2]^2)/(8*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d)
- (11*Log[Cos[(c + d*x)/2]])/(2*d) + (11*Log[Sin[(c + d*x)/2]])/(2*d) + (3*Sec[(c + d*x)/2]^2)/(8*d) + (8*Sin
[(c + d*x)/2])/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (7*Tan[(c + d*x)/2])/(3*d) + (Sec[(c + d*x)/2]^2*Ta
n[(c + d*x)/2])/(24*d))

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fricas [B]  time = 0.47, size = 354, normalized size = 3.61 \[ -\frac {104 \, a^{3} \cos \left (d x + c\right )^{4} + 38 \, a^{3} \cos \left (d x + c\right )^{3} - 156 \, a^{3} \cos \left (d x + c\right )^{2} - 42 \, a^{3} \cos \left (d x + c\right ) + 48 \, a^{3} + 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 33 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2} - a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (52 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} \cos \left (d x + c\right )^{2} - 45 \, a^{3} \cos \left (d x + c\right ) - 24 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(104*a^3*cos(d*x + c)^4 + 38*a^3*cos(d*x + c)^3 - 156*a^3*cos(d*x + c)^2 - 42*a^3*cos(d*x + c) + 48*a^3
+ 33*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3 + (a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2 - a^3*cos(d*x
+ c) - a^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 33*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3 +
(a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2 - a^3*cos(d*x + c) - a^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2)
- 2*(52*a^3*cos(d*x + c)^3 + 33*a^3*cos(d*x + c)^2 - 45*a^3*cos(d*x + c) - 24*a^3)*sin(d*x + c))/(d*cos(d*x +
c)^4 - 2*d*cos(d*x + c)^2 + (d*cos(d*x + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) - d)*sin(d*x + c) + d)

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giac [A]  time = 0.23, size = 148, normalized size = 1.51 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 132 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 57 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {192 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} - \frac {242 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 57 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c)^2 + 132*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 57*
a^3*tan(1/2*d*x + 1/2*c) - 192*a^3/(tan(1/2*d*x + 1/2*c) - 1) - (242*a^3*tan(1/2*d*x + 1/2*c)^3 + 57*a^3*tan(1
/2*d*x + 1/2*c)^2 + 9*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.59, size = 128, normalized size = 1.31 \[ \frac {11 a^{3}}{2 d \cos \left (d x +c \right )}+\frac {11 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {13 a^{3}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {26 a^{3} \cot \left (d x +c \right )}{3 d}-\frac {3 a^{3}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}-\frac {a^{3}}{3 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

11/2/d*a^3/cos(d*x+c)+11/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))+13/3/d*a^3/sin(d*x+c)/cos(d*x+c)-26/3*a^3*cot(d*x+c
)/d-3/2/d*a^3/sin(d*x+c)^2/cos(d*x+c)-1/3/d*a^3/sin(d*x+c)^3/cos(d*x+c)

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maxima [A]  time = 0.33, size = 160, normalized size = 1.63 \[ \frac {9 \, a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 36 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} - 4 \, a^{3} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(9*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)) - 36*a^3*(1/tan(d*x + c)
- tan(d*x + c)) - 4*a^3*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)))/d

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mupad [B]  time = 8.97, size = 160, normalized size = 1.63 \[ \frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {-83\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {8\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {11\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {19\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \]

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)^4),x)

[Out]

(3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - (16*a^3*tan(c/2 + (d*x)/2)^2 - 83*a^3
*tan(c/2 + (d*x)/2)^3 + a^3/3 + (8*a^3*tan(c/2 + (d*x)/2))/3)/(d*(8*tan(c/2 + (d*x)/2)^3 - 8*tan(c/2 + (d*x)/2
)^4)) + (11*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) + (19*a^3*tan(c/2 + (d*x)/2))/(8*d)

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

[Out]

Timed out

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