3.1461 \(\int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

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

[Out]

(-3*a^2*b*ArcTanh[Cos[c + d*x]])/d - (a^3*Cot[c + d*x])/d + (3*a^2*b*Sec[c + d*x])/d + (b^3*Sec[c + d*x])/d +
(a^3*Tan[c + d*x])/d + (3*a*b^2*Tan[c + d*x])/d

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Rubi [A]  time = 0.195415, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2912, 3767, 8, 2622, 321, 207, 2620, 14, 2606} \[ \frac{3 a^2 b \sec (c+d x)}{d}-\frac{3 a^2 b \tanh ^{-1}(\cos (c+d x))}{d}+\frac{a^3 \tan (c+d x)}{d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \tan (c+d x)}{d}+\frac{b^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*b*ArcTanh[Cos[c + d*x]])/d - (a^3*Cot[c + d*x])/d + (3*a^2*b*Sec[c + d*x])/d + (b^3*Sec[c + d*x])/d +
(a^3*Tan[c + d*x])/d + (3*a*b^2*Tan[c + d*x])/d

Rule 2912

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] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

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 8

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

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 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 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 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 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 2606

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

Rubi steps

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

Mathematica [A]  time = 0.389728, size = 114, normalized size = 1.31 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (-2 b \left (3 a^2+b^2\right ) \sin (c+d x)+\left (2 a^3+3 a b^2\right ) \cos (2 (c+d x))-3 a b \left (a \sin (2 (c+d x)) \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+b\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*Sec[c + d*x]*((2*a^3 + 3*a*b^2)*Cos[2*(c + d*x)] - 2*b*(3*a^2 + b^2)*Sin[c
+ d*x] - 3*a*b*(b + a*(-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]])*Sin[2*(c + d*x)])))/(4*d)

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Maple [A]  time = 0.082, size = 111, normalized size = 1.3 \begin{align*}{\frac{{a}^{3}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-2\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b}{d\cos \left ( dx+c \right ) }}+3\,{\frac{{a}^{2}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}}{d\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00241, size = 122, normalized size = 1.4 \begin{align*} \frac{3 \, a^{2} b{\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 b^{2} \tan \left (d x + c\right ) + \frac{2 \, b^{3}}{\cos \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a^2*b*(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*b^2*tan(d*x + c) + 2*b^3/cos(d*x + c))/d

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Fricas [A]  time = 1.66043, size = 342, normalized size = 3.93 \begin{align*} -\frac{3 \, a^{2} b \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \, a^{2} b \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a^{3} - 6 \, a b^{2} + 2 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.2468, size = 200, normalized size = 2.3 \begin{align*} \frac{6 \, a^{2} b \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{2 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 10 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, b^{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} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + a^3*tan(1/2*d*x + 1/2*c) - (2*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 5*a
^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a^2*b*tan(1/2*d*x + 1/2*c) + 4*b^3*tan(1/2*d*
x + 1/2*c) - a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)))/d