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

Optimal. Leaf size=56 \[ -\frac {a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2872, 3770, 3767, 8, 2648} \[ -\frac {a^3 \cot (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.37, size = 96, normalized size = 1.71 \[ \frac {a^3 \left (\tan \left (\frac {1}{2} (c+d x)\right )-\cot \left (\frac {1}{2} (c+d x)\right )+6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {16 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-Cot[(c + d*x)/2] - 6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]] + (16*Sin[(c + d*x)/2])/(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]) + Tan[(c + d*x)/2]))/(2*d)

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fricas [B]  time = 0.46, size = 194, normalized size = 3.46 \[ -\frac {10 \, a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \cos \left (d x + c\right ) - 8 \, a^{3} + 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} + {\left (a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3} + {\left (a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (5 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(10*a^3*cos(d*x + c)^2 + 2*a^3*cos(d*x + c) - 8*a^3 + 3*(a^3*cos(d*x + c)^2 - a^3 + (a^3*cos(d*x + c) + a
^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^3*cos(d*x + c)^2 - a^3 + (a^3*cos(d*x + c) + a^3)*sin(d*x
+ c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(5*a^3*cos(d*x + c) + 4*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 + (d*cos(
d*x + c) + d)*sin(d*x + c) - d)

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giac [A]  time = 0.21, size = 98, normalized size = 1.75 \[ \frac {6 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + a^3*tan(1/2*d*x + 1/2*c) - (3*a^3*tan(1/2*d*x + 1/2*c)^2 + 14*a^3*
tan(1/2*d*x + 1/2*c) - a^3)/(tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c)))/d

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maple [A]  time = 0.61, size = 93, normalized size = 1.66 \[ \frac {4 a^{3}}{d \cos \left (d x +c \right )}+\frac {3 a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {a^{3}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {2 a^{3} \cot \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 88, normalized size = 1.57 \[ \frac {3 \, a^{3} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + 6 \, a^{3} \tan \left (d x + c\right ) + \frac {2 \, a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.97, size = 86, normalized size = 1.54 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^3-17\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

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)^2),x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d - (a^3 - 17*a^3*tan(c/2 + (d*x)/2))/(d*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (
d*x)/2)^2)) + (a^3*tan(c/2 + (d*x)/2))/(2*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)**2*sec(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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