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]

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

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Rubi [A]  time = 0.29, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 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 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 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 ^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*}

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Mathematica [B]  time = 0.63, 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]

-1/2*(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*Co
s[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)*Lo
g[Cos[(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*Log[Sin[(c + d*x)/2]])))/(d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))

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fricas [A]  time = 0.54, size = 205, normalized size = 1.80 \[ \frac {4 \, a b \cos \left (d x + c\right ) + 2 \, {\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, b^{2} - 8 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right )\right ) - {\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )}} \]

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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giac [B]  time = 0.29, size = 314, normalized size = 2.75 \[ -\frac {16 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac {a^{2} + 2 \, a b + b^{2} + \frac {6 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {14 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {3 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac {{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{8 \, d} \]

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

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maple [A]  time = 0.74, size = 139, normalized size = 1.22 \[ -\frac {a^{2} \cot \left (d x +c \right ) \csc \left (d x +c \right )}{2 d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {a b}{d \sin \left (d x +c \right )^{2}}+\frac {2 a b \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {b^{2}}{2 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )}+\frac {3 b^{2}}{2 d \cos \left (d x +c \right )}+\frac {3 b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

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*a^2*cot(d*x+c)*csc(d*x+c)/d+1/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-1/d*a*b/sin(d*x+c)^2+2*a*b*ln(tan(d*x+c))
/d-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 = 0.67, size = 119, normalized size = 1.04 \[ -\frac {8 \, a b \log \left (\cos \left (d x + c\right )\right ) + {\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (2 \, a b \cos \left (d x + c\right ) + {\left (a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )}}{4 \, d} \]

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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mupad [B]  time = 0.12, size = 120, normalized size = 1.05 \[ \frac {\ln \left (\cos \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (a+3\,b\right )}{4\,d}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (a-3\,b\right )}{4\,d}-\frac {2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}+\frac {3\,b^2}{2}\right )-b^2+a\,b\,\cos \left (c+d\,x\right )}{d\,\left (\cos \left (c+d\,x\right )-{\cos \left (c+d\,x\right )}^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^2/sin(c + d*x)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \csc ^{3}{\left (c + d x \right )}\, dx \]

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]

Integral((a + b*sec(c + d*x))**2*csc(c + d*x)**3, x)

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