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

Optimal. Leaf size=74 \[ -\frac {(a-b)^2 \log (\cos (c+d x)+1)}{2 d}+\frac {(a+b)^2 \log (1-\cos (c+d x))}{2 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3872, 2837, 12, 1802} \[ -\frac {(a-b)^2 \log (\cos (c+d x)+1)}{2 d}+\frac {(a+b)^2 \log (1-\cos (c+d x))}{2 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 91, normalized size = 1.23 \[ \frac {a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (\cos (c+d x))-(a-b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+b^2 \sec (c+d x)+b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 97, normalized size = 1.31 \[ -\frac {4 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, b^{2}}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.51, size = 124, normalized size = 1.68 \[ -\frac {4 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac {4 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.38, size = 77, normalized size = 1.04 \[ \frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {2 a b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b^{2}}{d \cos \left (d x +c \right )}+\frac {b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.74, size = 73, normalized size = 0.99 \[ -\frac {4 \, a b \log \left (\cos \left (d x + c\right )\right ) + {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, b^{2}}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.97, size = 62, normalized size = 0.84 \[ \frac {\frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2}{2}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^2}{2}+\frac {b^2}{\cos \left (c+d\,x\right )}-2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^2/sin(c + d*x),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*csc(c + d*x), x)

________________________________________________________________________________________