3.1462 \(\int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=132 \[ \frac {3 a^3 \sec (c+d x)}{2 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {3 a^2 b \tan (c+d x)}{d}-\frac {3 a^2 b \cot (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {b^3 \tan (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2912, 3767, 8, 2622, 321, 207, 2620, 14, 288} \[ \frac {3 a^2 b \tan (c+d x)}{d}-\frac {3 a^2 b \cot (c+d x)}{d}+\frac {3 a^3 \sec (c+d x)}{2 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {b^3 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.57, size = 267, normalized size = 2.02 \[ \frac {\csc ^4(c+d x) \left (-6 \left (a^3+2 a b^2\right ) \cos (2 (c+d x))+3 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^3-3 a \left (a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+12 a^2 b \sin (c+d x)-12 a^2 b \sin (3 (c+d x))+6 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a b^2+6 b^3 \sin (c+d x)-2 b^3 \sin (3 (c+d x))\right )}{2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]^4*(2*a^3 + 12*a*b^2 - 6*(a^3 + 2*a*b^2)*Cos[2*(c + d*x)] + 3*a^3*Cos[3*(c + d*x)]*Log[Cos[(c + d
*x)/2]] + 6*a*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*a*(a^2 + 2*b^2)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2
]] - Log[Sin[(c + d*x)/2]]) - 3*a^3*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 6*a*b^2*Cos[3*(c + d*x)]*Log[Sin[
(c + d*x)/2]] + 12*a^2*b*Sin[c + d*x] + 6*b^3*Sin[c + d*x] - 12*a^2*b*Sin[3*(c + d*x)] - 2*b^3*Sin[3*(c + d*x)
]))/(2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 196, normalized size = 1.48 \[ -\frac {4 \, a^{3} + 12 \, a b^{2} - 6 \, {\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, {\left (3 \, a^{2} b + b^{3} - {\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.26, size = 179, normalized size = 1.36 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, {\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {16 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {18 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c) + 12*(a^3 + 2*a*b^2)*log(abs(tan(1/2*d*x + 1/2
*c))) - 16*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) + a^3 + 3*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 -
1) - (18*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/
tan(1/2*d*x + 1/2*c)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.73, size = 161, normalized size = 1.22 \[ -\frac {a^{3}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3 a^{3}}{2 d \cos \left (d x +c \right )}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}+\frac {3 a^{2} b}{d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {6 a^{2} b \cot \left (d x +c \right )}{d}+\frac {3 a \,b^{2}}{d \cos \left (d x +c \right )}+\frac {3 a \,b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {b^{3} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 137, normalized size = 1.04 \[ \frac {a^{3} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a b^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} b {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + 4 \, b^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2*(a+b*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*b^2*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a^2*b*(1/tan(d*x + c
) - tan(d*x + c)) + 4*b^3*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 11.91, size = 166, normalized size = 1.26 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^3}{2}+3\,a\,b^2\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {17\,a^3}{2}+24\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (30\,a^2\,b+8\,b^3\right )-\frac {a^3}{2}-6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*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) + (log(tan(c/2 + (d*x)/2))*(3*a*b^2 + (3*a^3)/2))/d + (tan(c/2 + (d*x)/2)^2*(
24*a*b^2 + (17*a^3)/2) + tan(c/2 + (d*x)/2)^3*(30*a^2*b + 8*b^3) - a^3/2 - 6*a^2*b*tan(c/2 + (d*x)/2))/(d*(4*t
an(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^4)) + (3*a^2*b*tan(c/2 + (d*x)/2))/(2*d)

________________________________________________________________________________________

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+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________