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

Optimal. Leaf size=86 \[ \frac {2 a^2 \tan (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {5 a^2 \sec (c+d x)}{2 d}-\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

[Out]

-5/2*a^2*arctanh(cos(d*x+c))/d-2*a^2*cot(d*x+c)/d+5/2*a^2*sec(d*x+c)/d-1/2*a^2*csc(d*x+c)^2*sec(d*x+c)/d+2*a^2
*tan(d*x+c)/d

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Rubi [A]  time = 0.21, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2622, 321, 207, 2620, 14, 288} \[ \frac {2 a^2 \tan (c+d x)}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {5 a^2 \sec (c+d x)}{2 d}-\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^2*ArcTanh[Cos[c + d*x]])/(2*d) - (2*a^2*Cot[c + d*x])/d + (5*a^2*Sec[c + d*x])/(2*d) - (a^2*Csc[c + d*x]
^2*Sec[c + d*x])/(2*d) + (2*a^2*Tan[c + d*x])/d

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2873

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

Rubi steps

\begin {align*} \int \csc ^3(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \csc (c+d x) \sec ^2(c+d x)+2 a^2 \csc ^2(c+d x) \sec ^2(c+d x)+a^2 \csc ^3(c+d x) \sec ^2(c+d x)\right ) \, dx\\ &=a^2 \int \csc (c+d x) \sec ^2(c+d x) \, dx+a^2 \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx\\ &=\frac {a^2 \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a^2 \sec (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {5 a^2 \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {2 a^2 \cot (c+d x)}{d}+\frac {5 a^2 \sec (c+d x)}{2 d}-\frac {a^2 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 124, normalized size = 1.44 \[ \frac {a^2 \left (8 \tan \left (\frac {1}{2} (c+d x)\right )-8 \cot \left (\frac {1}{2} (c+d x)\right )-\csc ^2\left (\frac {1}{2} (c+d x)\right )+\sec ^2\left (\frac {1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-20 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {32 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 20*Log[Cos[(c + d*x)/2]] + 20*Log[Sin[(c + d*x)/2]] + Sec[(c
+ d*x)/2]^2 + (32*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 8*Tan[(c + d*x)/2]))/(8*d)

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fricas [B]  time = 0.45, size = 300, normalized size = 3.49 \[ \frac {16 \, a^{2} \cos \left (d x + c\right )^{3} + 10 \, a^{2} \cos \left (d x + c\right )^{2} - 14 \, a^{2} \cos \left (d x + c\right ) - 8 \, a^{2} - 5 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (8 \, a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/4*(16*a^2*cos(d*x + c)^3 + 10*a^2*cos(d*x + c)^2 - 14*a^2*cos(d*x + c) - 8*a^2 - 5*(a^2*cos(d*x + c)^3 + a^2
*cos(d*x + c)^2 - a^2*cos(d*x + c) - a^2 - (a^2*cos(d*x + c)^2 - a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2
) + 5*(a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - a^2 - (a^2*cos(d*x + c)^2 - a^2)*sin(d*x +
c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(8*a^2*cos(d*x + c)^2 + 3*a^2*cos(d*x + c) - 4*a^2)*sin(d*x + c))/(d*cos
(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 = 1.35 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {32 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} - \frac {30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 20*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 8*a^2*tan(1/2*d*x + 1/2*c) - 32*a^2/
(tan(1/2*d*x + 1/2*c) - 1) - (30*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x +
1/2*c)^2)/d

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maple [A]  time = 0.60, size = 104, normalized size = 1.21 \[ \frac {5 a^{2}}{2 d \cos \left (d x +c \right )}+\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2}}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {4 a^{2} \cot \left (d x +c \right )}{d}-\frac {a^{2}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )} \]

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))^2,x)

[Out]

5/2/d*a^2/cos(d*x+c)+5/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))+2/d*a^2/sin(d*x+c)/cos(d*x+c)-4*a^2*cot(d*x+c)/d-1/2/
d*a^2/sin(d*x+c)^2/cos(d*x+c)

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maxima [A]  time = 0.33, size = 124, normalized size = 1.44 \[ \frac {a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{4 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

1/4*(a^2*(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)) + 2*a^2*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 8*a^2*(1/tan(d*x + c) - t
an(d*x + c)))/d

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mupad [B]  time = 8.96, size = 124, normalized size = 1.44 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a^2}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (5*a^2*log(tan(c/2 + (d*x)/2)))/(2*d) + (a^2*tan(c/2 + (d*x)/2))/d - (a^2/2
- 20*a^2*tan(c/2 + (d*x)/2)^2 + (7*a^2*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))**2,x)

[Out]

Timed out

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