3.1449 \(\int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=75 \[ \frac{3 a \sec (c+d x)}{2 d}-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{b \tan (c+d x)}{d}-\frac{b \cot (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.136787, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2838, 2622, 288, 321, 207, 2620, 14} \[ \frac{3 a \sec (c+d x)}{2 d}-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{b \tan (c+d x)}{d}-\frac{b \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, 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 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 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]]

Rubi steps

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

Mathematica [B]  time = 0.337545, size = 172, normalized size = 2.29 \[ -\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{3 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{3 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{a \sin \left (\frac{1}{2} (c+d x)\right )}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{a \sin \left (\frac{1}{2} (c+d x)\right )}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 b \cot (2 (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Cot[2*(c + d*x)])/d - (a*Csc[(c + d*x)/2]^2)/(8*d) - (3*a*Log[Cos[(c + d*x)/2]])/(2*d) + (3*a*Log[Sin[(c
+ d*x)/2]])/(2*d) + (a*Sec[(c + d*x)/2]^2)/(8*d) + (a*Sin[(c + d*x)/2])/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/
2])) - (a*Sin[(c + d*x)/2])/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.06, size = 93, normalized size = 1.2 \begin{align*} -{\frac{a}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{3\,a}{2\,d\cos \left ( dx+c \right ) }}+{\frac{3\,a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{b}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-2\,{\frac{b\cot \left ( dx+c \right ) }{d}} \end{align*}

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)),x)

[Out]

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

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Maxima [A]  time = 0.986827, size = 113, normalized size = 1.51 \begin{align*} \frac{a{\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 )} - 4 \, b{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{4 \, d} \end{align*}

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)),x, algorithm="maxima")

[Out]

1/4*(a*(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)) - 4*b*(1/tan(d*x + c) - tan(d*x + c)))/d

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Fricas [A]  time = 1.66725, size = 333, normalized size = 4.44 \begin{align*} \frac{6 \, a \cos \left (d x + c\right )^{2} - 3 \,{\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 4 \,{\left (2 \, b \cos \left (d x + c\right )^{2} - b\right )} \sin \left (d x + c\right ) - 4 \, a}{4 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )}} \end{align*}

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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2304, size = 157, normalized size = 2.09 \begin{align*} \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{16 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{18 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

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)),x, algorithm="giac")

[Out]

1/8*(a*tan(1/2*d*x + 1/2*c)^2 + 12*a*log(abs(tan(1/2*d*x + 1/2*c))) + 4*b*tan(1/2*d*x + 1/2*c) - 16*(b*tan(1/2
*d*x + 1/2*c) + a)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (18*a*tan(1/2*d*x + 1/2*c)^2 + 4*b*tan(1/2*d*x + 1/2*c) + a)
/tan(1/2*d*x + 1/2*c)^2)/d