3.1488 \(\int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=99 \[ \frac{a \tan ^4(c+d x)}{4 d}+\frac{a \tan ^2(c+d x)}{d}+\frac{a \log (\tan (c+d x))}{d}+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

(3*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a*Log[Tan[c + d*x]])/d + (3*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c
+ d*x]^3*Tan[c + d*x])/(4*d) + (a*Tan[c + d*x]^2)/d + (a*Tan[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.105285, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2834, 2620, 266, 43, 3768, 3770} \[ \frac{a \tan ^4(c+d x)}{4 d}+\frac{a \tan ^2(c+d x)}{d}+\frac{a \log (\tan (c+d x))}{d}+\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

(3*b*ArcTanh[Sin[c + d*x]])/(8*d) + (a*Log[Tan[c + d*x]])/d + (3*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c
+ d*x]^3*Tan[c + d*x])/(4*d) + (a*Tan[c + d*x]^2)/d + (a*Tan[c + d*x]^4)/(4*d)

Rule 2834

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \csc (c+d x) \sec ^5(c+d x) \, dx+b \int \sec ^5(c+d x) \, dx\\ &=\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} (3 b) \int \sec ^3(c+d x) \, dx+\frac{a \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} (3 b) \int \sec (c+d x) \, dx+\frac{a \operatorname{Subst}\left (\int \frac{(1+x)^2}{x} \, dx,x,\tan ^2(c+d x)\right )}{2 d}\\ &=\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a \operatorname{Subst}\left (\int \left (2+\frac{1}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d}\\ &=\frac{3 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \log (\tan (c+d x))}{d}+\frac{3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac{b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a \tan ^2(c+d x)}{d}+\frac{a \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.25655, size = 99, normalized size = 1. \[ -\frac{a \left (-\sec ^4(c+d x)-2 \sec ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d}+\frac{b \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{3 b \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]

[Out]

-(a*(4*Log[Cos[c + d*x]] - 4*Log[Sin[c + d*x]] - 2*Sec[c + d*x]^2 - Sec[c + d*x]^4))/(4*d) + (b*Sec[c + d*x]^3
*Tan[c + d*x])/(4*d) + (3*b*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x]))/(8*d)

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Maple [A]  time = 0.059, size = 100, normalized size = 1. \begin{align*}{\frac{a}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{a}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+{\frac{b \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,b\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x)

[Out]

1/4/d*a/cos(d*x+c)^4+1/2/d*a/cos(d*x+c)^2+a*ln(tan(d*x+c))/d+1/4*b*sec(d*x+c)^3*tan(d*x+c)/d+3/8*b*sec(d*x+c)*
tan(d*x+c)/d+3/8/d*b*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.999947, size = 147, normalized size = 1.48 \begin{align*} -\frac{{\left (8 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (8 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac{2 \,{\left (3 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 5 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*((8*a - 3*b)*log(sin(d*x + c) + 1) + (8*a + 3*b)*log(sin(d*x + c) - 1) - 16*a*log(sin(d*x + c)) + 2*(3*b
*sin(d*x + c)^3 + 4*a*sin(d*x + c)^2 - 5*b*sin(d*x + c) - 6*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 2.04968, size = 328, normalized size = 3.31 \begin{align*} \frac{16 \, a \cos \left (d x + c\right )^{4} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) -{\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(16*a*cos(d*x + c)^4*log(1/2*sin(d*x + c)) - (8*a - 3*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a + 3*
b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*a*cos(d*x + c)^2 + 2*(3*b*cos(d*x + c)^2 + 2*b)*sin(d*x + c) + 4*
a)/(d*cos(d*x + c)^4)

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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)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.223, size = 153, normalized size = 1.55 \begin{align*} -\frac{{\left (8 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (8 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{2 \,{\left (6 \, a \sin \left (d x + c\right )^{4} - 3 \, b \sin \left (d x + c\right )^{3} - 16 \, a \sin \left (d x + c\right )^{2} + 5 \, b \sin \left (d x + c\right ) + 12 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*((8*a - 3*b)*log(abs(sin(d*x + c) + 1)) + (8*a + 3*b)*log(abs(sin(d*x + c) - 1)) - 16*a*log(abs(sin(d*x
+ c))) - 2*(6*a*sin(d*x + c)^4 - 3*b*sin(d*x + c)^3 - 16*a*sin(d*x + c)^2 + 5*b*sin(d*x + c) + 12*a)/(sin(d*x
+ c)^2 - 1)^2)/d