3.188 \(\int \csc (c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=102 \[ -\frac{b \left (3 a^2+b^2\right ) \log (\cos (c+d x))}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{(a-b)^3 \log (\cos (c+d x)+1)}{2 d}+\frac{(a+b)^3 \log (1-\cos (c+d x))}{2 d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

[Out]

((a + b)^3*Log[1 - Cos[c + d*x]])/(2*d) - (b*(3*a^2 + b^2)*Log[Cos[c + d*x]])/d - ((a - b)^3*Log[1 + Cos[c + d
*x]])/(2*d) + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.219213, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3872, 2837, 12, 1802} \[ -\frac{b \left (3 a^2+b^2\right ) \log (\cos (c+d x))}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{(a-b)^3 \log (\cos (c+d x)+1)}{2 d}+\frac{(a+b)^3 \log (1-\cos (c+d x))}{2 d}+\frac{b^3 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^3*Log[1 - Cos[c + d*x]])/(2*d) - (b*(3*a^2 + b^2)*Log[Cos[c + d*x]])/d - ((a - b)^3*Log[1 + Cos[c + d
*x]])/(2*d) + (3*a*b^2*Sec[c + d*x])/d + (b^3*Sec[c + d*x]^2)/(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 2837

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.292273, size = 89, normalized size = 0.87 \[ \frac{-2 b \left (3 a^2+b^2\right ) \log (\cos (c+d x))+6 a b^2 \sec (c+d x)+2 (a+b)^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 (a-b)^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+b^3 \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)^3*Log[Cos[(c + d*x)/2]] - 2*b*(3*a^2 + b^2)*Log[Cos[c + d*x]] + 2*(a + b)^3*Log[Sin[(c + d*x)/2]]
+ 6*a*b^2*Sec[c + d*x] + b^3*Sec[c + d*x]^2)/(2*d)

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Maple [A]  time = 0.041, size = 113, normalized size = 1.1 \begin{align*}{\frac{{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{2}b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{a{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.967473, size = 151, normalized size = 1.48 \begin{align*} -\frac{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) -{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) + 2 \,{\left (3 \, a^{2} b + b^{3}\right )} \log \left (\cos \left (d x + c\right )\right ) - \frac{6 \, a b^{2} \cos \left (d x + c\right ) + b^{3}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93014, size = 354, normalized size = 3.47 \begin{align*} \frac{6 \, a b^{2} \cos \left (d x + c\right ) - 2 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) -{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + b^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{3} \csc{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*sec(d*x+c))**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*csc(c + d*x), x)

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Giac [B]  time = 1.49241, size = 338, normalized size = 3.31 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 2 \,{\left (3 \, a^{2} b + b^{3}\right )} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{9 \, a^{2} b + 12 \, a b^{2} + 3 \, b^{3} + \frac{18 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{12 \, a b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{2 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \, a^{2} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 2*(3*a^2*b + b^3)*log
(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + (9*a^2*b + 12*a*b^2 + 3*b^3 + 18*a^2*b*(cos(d*x + c) - 1)/
(cos(d*x + c) + 1) + 12*a*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 9*a^2*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d