3.201 \(\int \frac{\csc ^3(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=116 \[ \frac{a^2 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^2}+\frac{\csc ^2(c+d x) (b-a \cos (c+d x))}{2 d \left (a^2-b^2\right )}+\frac{a \log (1-\cos (c+d x))}{4 d (a+b)^2}-\frac{a \log (\cos (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

((b - a*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d) + (a*Log[1 - Cos[c + d*x]])/(4*(a + b)^2*d) - (a*Log[1
+ Cos[c + d*x]])/(4*(a - b)^2*d) + (a^2*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^2)^2*d)

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Rubi [A]  time = 0.21299, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2837, 12, 823, 801} \[ \frac{a^2 b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^2}+\frac{\csc ^2(c+d x) (b-a \cos (c+d x))}{2 d \left (a^2-b^2\right )}+\frac{a \log (1-\cos (c+d x))}{4 d (a+b)^2}-\frac{a \log (\cos (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b - a*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(a^2 - b^2)*d) + (a*Log[1 - Cos[c + d*x]])/(4*(a + b)^2*d) - (a*Log[1
+ Cos[c + d*x]])/(4*(a - b)^2*d) + (a^2*b*Log[b + a*Cos[c + d*x]])/((a^2 - b^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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.595094, size = 123, normalized size = 1.06 \[ \frac{-(a-b)^2 (a+b) \csc ^2\left (\frac{1}{2} (c+d x)\right )+(a-b) (a+b)^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )-4 a \left ((a-b)^2 \left (-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+(a+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 a b \log (a \cos (c+d x)+b)\right )}{8 d (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.069, size = 121, normalized size = 1. \begin{align*}{\frac{{a}^{2}b\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}+{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{4\, \left ( a-b \right ) ^{2}d}}+{\frac{1}{d \left ( 4\,a+4\,b \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{a\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{4\,d \left ( a+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*b*a^2/(a+b)^2/(a-b)^2*ln(b+a*cos(d*x+c))+1/d/(4*a-4*b)/(cos(d*x+c)+1)-1/4*a*ln(cos(d*x+c)+1)/(a-b)^2/d+1/d
/(4*a+4*b)/(-1+cos(d*x+c))+1/4/d*a/(a+b)^2*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.00064, size = 178, normalized size = 1.53 \begin{align*} \frac{\frac{4 \, a^{2} b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{a \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{a \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (a \cos \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.17163, size = 508, normalized size = 4.38 \begin{align*} -\frac{2 \, a^{2} b - 2 \, b^{3} - 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) - 4 \,{\left (a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (a^{3} + 2 \, a^{2} b + a b^{2} -{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{3} - 2 \, a^{2} b + a b^{2} -{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32966, size = 273, normalized size = 2.35 \begin{align*} \frac{\frac{8 \, a^{2} b \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \, a \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{{\left (a + b - \frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{\cos \left (d x + c\right ) - 1}{{\left (a - b\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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