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

Optimal. Leaf size=114 \[ -\frac{\csc ^2(c+d x) \left (\left (a^2+b^2\right ) \cos (c+d x)+2 a b\right )}{2 d}-\frac{2 a b \log (\cos (c+d x))}{d}+\frac{(a+b) (a+3 b) \log (1-\cos (c+d x))}{4 d}-\frac{(a-3 b) (a-b) \log (\cos (c+d x)+1)}{4 d}+\frac{b^2 \sec (c+d x)}{d} \]

[Out]

-((2*a*b + (a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*d) + ((a + b)*(a + 3*b)*Log[1 - Cos[c + d*x]])/(4*d) -
(2*a*b*Log[Cos[c + d*x]])/d - ((a - 3*b)*(a - b)*Log[1 + Cos[c + d*x]])/(4*d) + (b^2*Sec[c + d*x])/d

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Rubi [A]  time = 0.294116, antiderivative size = 114, 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, 1805, 1802} \[ -\frac{\csc ^2(c+d x) \left (\left (a^2+b^2\right ) \cos (c+d x)+2 a b\right )}{2 d}-\frac{2 a b \log (\cos (c+d x))}{d}+\frac{(a+b) (a+3 b) \log (1-\cos (c+d x))}{4 d}-\frac{(a-3 b) (a-b) \log (\cos (c+d x)+1)}{4 d}+\frac{b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*b + (a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x]^2)/(2*d) + ((a + b)*(a + 3*b)*Log[1 - Cos[c + d*x]])/(4*d) -
(2*a*b*Log[Cos[c + d*x]])/d - ((a - 3*b)*(a - b)*Log[1 + Cos[c + d*x]])/(4*d) + (b^2*Sec[c + d*x])/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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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

Mathematica [B]  time = 0.615619, size = 329, normalized size = 2.89 \[ -\frac{\csc ^4(c+d x) \left (2 \left (a^2+3 b^2\right ) \cos (2 (c+d x))+\cos (c+d x) \left (\left (a^2-4 a b+3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+a^2 \left (-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )-4 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 a b \log (\cos (c+d x))+8 a b-3 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+a^2 (-\cos (3 (c+d x))) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+a^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2+4 a b \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-4 a b \cos (3 (c+d x)) \log (\cos (c+d x))+4 a b \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 b^2\right )}{2 d \left (\csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^4*(2*a^2 - 2*b^2 + 2*(a^2 + 3*b^2)*Cos[2*(c + d*x)] - a^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]
] + 4*a*b*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*b^2*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 4*a*b*Cos[3*
(c + d*x)]*Log[Cos[c + d*x]] + a^2*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 4*a*b*Cos[3*(c + d*x)]*Log[Sin[(c
+ d*x)/2]] + 3*b^2*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(8*a*b + (a^2 - 4*a*b + 3*b^2)*Log[Co
s[(c + d*x)/2]] + 4*a*b*Log[Cos[c + d*x]] - a^2*Log[Sin[(c + d*x)/2]] - 4*a*b*Log[Sin[(c + d*x)/2]] - 3*b^2*Lo
g[Sin[(c + d*x)/2]])))/(2*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

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Maple [A]  time = 0.042, size = 139, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}\csc \left ( dx+c \right ) \cot \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{ab}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{ab\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{3\,{b}^{2}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{3\,{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 0.958066, size = 161, normalized size = 1.41 \begin{align*} -\frac{8 \, a b \log \left (\cos \left (d x + c\right )\right ) +{\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) -{\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (2 \, a b \cos \left (d x + c\right ) +{\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (d x + c\right )^{3} - \cos \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)^3*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.86678, size = 512, normalized size = 4.49 \begin{align*} \frac{4 \, a b \cos \left (d x + c\right ) + 2 \,{\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, b^{2} - 8 \,{\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) -{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left ({\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (d \cos \left (d x + c\right )^{3} - 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)^3*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(4*a*b*cos(d*x + c) + 2*(a^2 + 3*b^2)*cos(d*x + c)^2 - 4*b^2 - 8*(a*b*cos(d*x + c)^3 - a*b*cos(d*x + c))*l
og(-cos(d*x + c)) - ((a^2 - 4*a*b + 3*b^2)*cos(d*x + c)^3 - (a^2 - 4*a*b + 3*b^2)*cos(d*x + c))*log(1/2*cos(d*
x + c) + 1/2) + ((a^2 + 4*a*b + 3*b^2)*cos(d*x + c)^3 - (a^2 + 4*a*b + 3*b^2)*cos(d*x + c))*log(-1/2*cos(d*x +
c) + 1/2))/(d*cos(d*x + c)^3 - 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)**3*(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.38369, size = 424, normalized size = 3.72 \begin{align*} -\frac{16 \, a b \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \,{\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac{a^{2} + 2 \, a b + b^{2} + \frac{6 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{14 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{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))^2,x, algorithm="giac")

[Out]

-1/8*(16*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
- 2*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*(a^2 + 4*a*b + 3
*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - (a^2 + 2*a*b + b^2 + 6*a*b*(cos(d*x + c) - 1)/(cos(d
*x + c) + 1) + 14*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 +
4*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((cos(d*x +
c) - 1)/(cos(d*x + c) + 1) + (cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2))/d