3.183 \(\int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=100 \[ -\frac {\left (a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3872, 2911, 2621, 302, 207, 448} \[ -\frac {\left (a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

(2*a*b*ArcTanh[Sin[c + d*x]])/d - ((a^2 + 2*b^2)*Cot[c + d*x])/d - ((a^2 + b^2)*Cot[c + d*x]^3)/(3*d) - (2*a*b
*Csc[c + d*x])/d - (2*a*b*Csc[c + d*x]^3)/(3*d) + (b^2*Tan[c + d*x])/d

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && 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 ^4(c+d x) (a+b \sec (c+d x))^2 \, dx &=\int (-b-a \cos (c+d x))^2 \csc ^4(c+d x) \sec ^2(c+d x) \, dx\\ &=(2 a b) \int \csc ^4(c+d x) \sec (c+d x) \, dx+\int \left (b^2+a^2 \cos ^2(c+d x)\right ) \csc ^4(c+d x) \sec ^2(c+d x) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right ) \left (a^2+b^2+b^2 x^2\right )}{x^4} \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b^2+\frac {a^2+b^2}{x^4}+\frac {a^2+2 b^2}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac {\left (a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}+\frac {b^2 \tan (c+d x)}{d}-\frac {(2 a b) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac {\left (a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}+\frac {b^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 0.66, size = 259, normalized size = 2.59 \[ \frac {\csc ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-2 \left (a^2+4 b^2\right ) \cos (2 (c+d x))+a^2 \cos (4 (c+d x))-3 a^2-14 a b \cos (c+d x)+6 a b \cos (3 (c+d x))-6 a b \sin (2 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 a b \sin (2 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 a b \sin (4 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 a b \sin (4 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 \cos (4 (c+d x))\right )}{96 d \left (\cot ^2\left (\frac {1}{2} (c+d x)\right )-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]^3*(-3*a^2 - 14*a*b*Cos[c + d*x] - 2*(a^2 + 4*b^2)*Cos[2*(c + d*x)] + 6*a*
b*Cos[3*(c + d*x)] + a^2*Cos[4*(c + d*x)] + 4*b^2*Cos[4*(c + d*x)] - 6*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]]*Sin[2*(c + d*x)] + 6*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] + 3*a*b*Log[Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]]*Sin[4*(c + d*x)] - 3*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]
))/(96*d*(-1 + Cot[(c + d*x)/2]^2))

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fricas [A]  time = 0.55, size = 178, normalized size = 1.78 \[ -\frac {6 \, a b \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 8 \, a b \cos \left (d x + c\right ) - 3 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, b^{2}}{3 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.13, size = 226, normalized size = 2.26 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {48 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} + 2 \, a b + b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 2*a*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 + 48*a*b*log(abs(
tan(1/2*d*x + 1/2*c) + 1)) - 48*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 9*a^2*tan(1/2*d*x + 1/2*c) - 30*a*b*t
an(1/2*d*x + 1/2*c) + 21*b^2*tan(1/2*d*x + 1/2*c) - 48*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) -
(9*a^2*tan(1/2*d*x + 1/2*c)^2 + 30*a*b*tan(1/2*d*x + 1/2*c)^2 + 21*b^2*tan(1/2*d*x + 1/2*c)^2 + a^2 + 2*a*b +
b^2)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.92, size = 151, normalized size = 1.51 \[ -\frac {2 a^{2} \cot \left (d x +c \right )}{3 d}-\frac {a^{2} \cot \left (d x +c \right ) \left (\csc ^{2}\left (d x +c \right )\right )}{3 d}-\frac {2 a b}{3 d \sin \left (d x +c \right )^{3}}-\frac {2 a b}{d \sin \left (d x +c \right )}+\frac {2 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {b^{2}}{3 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4 b^{2}}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 b^{2} \cot \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*(a+b*sec(d*x+c))^2,x)

[Out]

-2/3*a^2*cot(d*x+c)/d-1/3/d*a^2*cot(d*x+c)*csc(d*x+c)^2-2/3/d*a*b/sin(d*x+c)^3-2/d*a*b/sin(d*x+c)+2/d*a*b*ln(s
ec(d*x+c)+tan(d*x+c))-1/3/d*b^2/sin(d*x+c)^3/cos(d*x+c)+4/3/d*b^2/sin(d*x+c)/cos(d*x+c)-8/3/d*b^2*cot(d*x+c)

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maxima [A]  time = 0.75, size = 112, normalized size = 1.12 \[ -\frac {a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + b^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.10, size = 182, normalized size = 1.82 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\left (a-b\right )}^2}{24\,d}-\frac {\frac {2\,a\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2+10\,a\,b+23\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {8\,a^2}{3}+\frac {28\,a\,b}{3}+\frac {20\,b^2}{3}\right )+\frac {a^2}{3}+\frac {b^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2}{8}-\frac {3\,a\,b}{4}+\frac {5\,b^2}{8}+\frac {{\left (a-b\right )}^2}{4}\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^2/sin(c + d*x)^4,x)

[Out]

(tan(c/2 + (d*x)/2)^3*(a - b)^2)/(24*d) - ((2*a*b)/3 - tan(c/2 + (d*x)/2)^4*(10*a*b + 3*a^2 + 23*b^2) + tan(c/
2 + (d*x)/2)^2*((28*a*b)/3 + (8*a^2)/3 + (20*b^2)/3) + a^2/3 + b^2/3)/(d*(8*tan(c/2 + (d*x)/2)^3 - 8*tan(c/2 +
(d*x)/2)^5)) + (tan(c/2 + (d*x)/2)*(a^2/8 - (3*a*b)/4 + (5*b^2)/8 + (a - b)^2/4))/d + (4*a*b*atanh(tan(c/2 +
(d*x)/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \csc ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**4*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*csc(c + d*x)**4, x)

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