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

Optimal. Leaf size=80 \[ \frac{3 a^3 \tan (c+d x)}{d}+\frac{9 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(9*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (4*a^3*Sin[c + d*x])/(d*(1 - Cos[c + d*x])) + (3*a^3*Tan[c + d*x])/d + (
a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.194048, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2872, 2648, 3770, 3767, 8, 3768} \[ \frac{3 a^3 \tan (c+d x)}{d}+\frac{9 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (4*a^3*Sin[c + d*x])/(d*(1 - Cos[c + d*x])) + (3*a^3*Tan[c + d*x])/d + (
a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2872

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3770

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

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps

\begin{align*} \int \csc ^2(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^2(c+d x) \sec ^3(c+d x) \, dx\\ &=a^2 \int \left (\frac{4 a}{1-\cos (c+d x)}+4 a \sec (c+d x)+3 a \sec ^2(c+d x)+a \sec ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (4 a^3\right ) \int \frac{1}{1-\cos (c+d x)} \, dx+\left (4 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac{4 a^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} a^3 \int \sec (c+d x) \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{9 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{4 a^3 \sin (c+d x)}{d (1-\cos (c+d x))}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 1.06539, size = 244, normalized size = 3.05 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{12 \sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+16 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc \left (\frac{1}{2} (c+d x)\right )-18 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+18 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(-18*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 18*Log[Cos[(c + d
*x)/2] + Sin[(c + d*x)/2]] + 16*Csc[c/2]*Csc[(c + d*x)/2]*Sin[(d*x)/2] + (Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
^(-2) - (Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^(-2) + (12*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]
)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(32*d)

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Maple [A]  time = 0.049, size = 102, normalized size = 1.3 \begin{align*} -7\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-{\frac{9\,{a}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{9\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}+{\frac{{a}^{3}}{2\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*(a+a*sec(d*x+c))^3,x)

[Out]

-7*a^3*cot(d*x+c)/d-9/2/d*a^3/sin(d*x+c)+9/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/d*a^3/sin(d*x+c)/cos(d*x+c)+1/2
/d*a^3/sin(d*x+c)/cos(d*x+c)^2

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Maxima [A]  time = 1.02052, size = 185, normalized size = 2.31 \begin{align*} -\frac{a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} - 2\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3}{\left (\frac{2}{\sin \left (d x + c\right )} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3}{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + \frac{4 \, a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*(a^3*(2*(3*sin(d*x + c)^2 - 2)/(sin(d*x + c)^3 - sin(d*x + c)) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x
+ c) - 1)) + 6*a^3*(2/sin(d*x + c) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*a^3*(1/tan(d*x + c) -
tan(d*x + c)) + 4*a^3/tan(d*x + c))/d

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Fricas [A]  time = 1.78353, size = 313, normalized size = 3.91 \begin{align*} \frac{9 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 9 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 28 \, a^{3} \cos \left (d x + c\right )^{3} - 18 \, a^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}}{4 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(9*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1)*sin(d*x + c) - 9*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1)*sin
(d*x + c) - 28*a^3*cos(d*x + c)^3 - 18*a^3*cos(d*x + c)^2 + 12*a^3*cos(d*x + c) + 2*a^3)/(d*cos(d*x + c)^2*sin
(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*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.30293, size = 143, normalized size = 1.79 \begin{align*} \frac{9 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{8 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(9*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 8*a^3/tan(1/2*d*x +
1/2*c) - 2*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d