3.758 \(\int \csc ^4(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=91 \[ \frac{a \tan (c+d x)}{d}-\frac{a \cot ^3(c+d x)}{3 d}-\frac{2 a \cot (c+d x)}{d}+\frac{3 a \sec (c+d x)}{2 d}-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.131479, antiderivative size = 91, 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, 2620, 270, 2622, 288, 321, 207} \[ \frac{a \tan (c+d x)}{d}-\frac{a \cot ^3(c+d x)}{3 d}-\frac{2 a \cot (c+d x)}{d}+\frac{3 a \sec (c+d x)}{2 d}-\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a \csc ^2(c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

(-3*a*ArcTanh[Cos[c + d*x]])/(2*d) - (2*a*Cot[c + d*x])/d - (a*Cot[c + d*x]^3)/(3*d) + (3*a*Sec[c + d*x])/(2*d
) - (a*Csc[c + d*x]^2*Sec[c + d*x])/(2*d) + (a*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 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 270

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

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])

Rubi steps

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

Mathematica [B]  time = 4.8328, size = 205, normalized size = 2.25 \[ \frac{a \tan (c+d x)}{d}-\frac{5 a \cot (c+d x)}{3 d}-\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{3 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{3 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{a \sin \left (\frac{1}{2} (c+d x)\right )}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{a \sin \left (\frac{1}{2} (c+d x)\right )}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{a \cot (c+d x) \csc ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^2*(a + a*Sin[c + d*x]),x]

[Out]

(-5*a*Cot[c + d*x])/(3*d) - (a*Csc[(c + d*x)/2]^2)/(8*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) - (3*a*Log[Co
s[(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]))
+ (a*Tan[c + d*x])/d

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Maple [A]  time = 0.084, size = 116, normalized size = 1.3 \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{a}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{4\,a}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{8\,a\cot \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*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))-1/3/d*a/sin(d*x+c)^3/cos
(d*x+c)+4/3/d*a/sin(d*x+c)/cos(d*x+c)-8/3*a*cot(d*x+c)/d

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Maxima [A]  time = 1.08284, size = 132, normalized size = 1.45 \begin{align*} \frac{3 \, 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 \, a{\left (\frac{6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(3*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*a*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)))/d

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Fricas [B]  time = 1.12838, size = 817, normalized size = 8.98 \begin{align*} -\frac{32 \, a \cos \left (d x + c\right )^{4} + 14 \, a \cos \left (d x + c\right )^{3} - 48 \, a \cos \left (d x + c\right )^{2} - 18 \, a \cos \left (d x + c\right ) + 9 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} +{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) + a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 9 \,{\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} +{\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) + a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (16 \, a \cos \left (d x + c\right )^{3} + 9 \, a \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 6 \, a\right )} \sin \left (d x + c\right ) + 12 \, a}{12 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} +{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(32*a*cos(d*x + c)^4 + 14*a*cos(d*x + c)^3 - 48*a*cos(d*x + c)^2 - 18*a*cos(d*x + c) + 9*(a*cos(d*x + c)
^4 - 2*a*cos(d*x + c)^2 + (a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) - a)*sin(d*x + c) + a)*log(1/2
*cos(d*x + c) + 1/2) - 9*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + (a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos
(d*x + c) - a)*sin(d*x + c) + a)*log(-1/2*cos(d*x + c) + 1/2) - 2*(16*a*cos(d*x + c)^3 + 9*a*cos(d*x + c)^2 -
15*a*cos(d*x + c) - 6*a)*sin(d*x + c) + 12*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + (d*cos(d*x + c)^3 + d*c
os(d*x + c)^2 - d*cos(d*x + c) - d)*sin(d*x + c) + d)

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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)**4*sec(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.27197, size = 176, normalized size = 1.93 \begin{align*} \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 21 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{48 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1} - \frac{66 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a \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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(a*tan(1/2*d*x + 1/2*c)^3 + 3*a*tan(1/2*d*x + 1/2*c)^2 + 36*a*log(abs(tan(1/2*d*x + 1/2*c))) + 21*a*tan(1
/2*d*x + 1/2*c) - 48*a/(tan(1/2*d*x + 1/2*c) - 1) - (66*a*tan(1/2*d*x + 1/2*c)^3 + 21*a*tan(1/2*d*x + 1/2*c)^2
+ 3*a*tan(1/2*d*x + 1/2*c) + a)/tan(1/2*d*x + 1/2*c)^3)/d