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

Optimal. Leaf size=68 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0874933, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2838, 2622, 302, 207, 3767} \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.116491, size = 85, normalized size = 1.25 \[ \frac{a \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{a \sec ^3(c+d x)}{3 d}+\frac{a \sec (c+d x)}{d}+\frac{a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}-\frac{a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.082, size = 82, normalized size = 1.2 \begin{align*}{\frac{2\,a\tan \left ( dx+c \right ) }{3\,d}}+{\frac{a\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x)

[Out]

2/3*a*tan(d*x+c)/d+1/3/d*a*tan(d*x+c)*sec(d*x+c)^2+1/3/d*a/cos(d*x+c)^3+1/d*a/cos(d*x+c)+1/d*a*ln(csc(d*x+c)-c
ot(d*x+c))

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Maxima [A]  time = 1.17048, size = 99, normalized size = 1.46 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \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)*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))*a + a*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x + c) +
1) + 3*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.67432, size = 348, normalized size = 5.12 \begin{align*} -\frac{4 \, a \cos \left (d x + c\right )^{2} + 3 \,{\left (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 ) \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 \, a \sin \left (d x + c\right ) + 4 \, a}{6 \,{\left (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)*sec(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.28876, size = 109, normalized size = 1.6 \begin{align*} \frac{6 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{3 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} - \frac{15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 13 \, 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)*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) + 1) - (15*a*tan(1/2*d*x + 1/2*c)^2 - 24*a
*tan(1/2*d*x + 1/2*c) + 13*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d