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

Optimal. Leaf size=81 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{2 a \tan (c+d x)}{d}-\frac{a \cot (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.123752, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2838, 2620, 270, 2622, 302, 207} \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{2 a \tan (c+d x)}{d}-\frac{a \cot (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTanh[Cos[c + d*x]])/d) - (a*Cot[c + d*x])/d + (a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) + (2*a*Tan
[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.0566675, size = 109, normalized size = 1.35 \[ \frac{5 a \tan (c+d x)}{3 d}-\frac{a \cot (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}+\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{a \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cot[c + d*x])/d) - (a*Log[Cos[(c + d*x)/2]])/d + (a*Log[Sin[(c + d*x)/2]])/d + (a*Sec[c + d*x])/d + (a*Se
c[c + d*x]^3)/(3*d) + (5*a*Tan[c + d*x])/(3*d) + (a*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Maple [A]  time = 0.098, size = 106, normalized size = 1.3 \begin{align*}{\frac{a}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a}{d\cos \left ( dx+c \right ) }}+{\frac{a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{a}{3\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\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)^2*sec(d*x+c)^4*(a+a*sin(d*x+c)),x)

[Out]

1/3/d*a/cos(d*x+c)^3+1/d*a/cos(d*x+c)+1/d*a*ln(csc(d*x+c)-cot(d*x+c))+1/3/d*a/sin(d*x+c)/cos(d*x+c)^3+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 = 1.01255, size = 112, normalized size = 1.38 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} - \frac{3}{\tan \left (d x + c\right )} + 6 \, \tan \left (d x + c\right )\right )} a + a{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(tan(d*x + c)^3 - 3/tan(d*x + c) + 6*tan(d*x + c))*a + a*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*l
og(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)))/d

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Fricas [B]  time = 1.61541, size = 458, normalized size = 5.65 \begin{align*} -\frac{10 \, 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 ) \sin \left (d x + c\right ) - 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 ) \sin \left (d x + c\right ) - a \cos \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 \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 4 \, a}{6 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 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)^2*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(10*a*cos(d*x + c)^2 + 3*(a*cos(d*x + c)^3 + a*cos(d*x + c)*sin(d*x + c) - 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)*sin(d*x + c) - a*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/
2) - 2*(8*a*cos(d*x + c)^2 - a)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^3 + d*cos(d*x + c)*sin(d*x + c) - 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)**2*sec(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.28096, size = 174, normalized size = 2.15 \begin{align*} \frac{6 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{3 \,{\left (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\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{21 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 19 \, a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a*tan(1/2*d*x + 1/2*c) - 3*(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)^2 + tan(1/2*d*x + 1/2*c)) - (21*a*tan(1/2*d*x + 1/2*c)^2 - 36*a*tan(1
/2*d*x + 1/2*c) + 19*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d