3.560 \(\int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}+\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d} \]

[Out]

-1/4*(2*a-b)*arctanh((a+b*sin(d*x+c)^4)^(1/2)/a^(1/2))/a^(3/2)/d+csc(d*x+c)^2*(a+b*sin(d*x+c)^4)^(1/2)/a/d-1/4
*csc(d*x+c)^4*(a+b*sin(d*x+c)^4)^(1/2)/a/d

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Rubi [A]  time = 0.18, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3229, 1807, 807, 266, 63, 208} \[ -\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}+\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-((2*a - b)*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]])/(4*a^(3/2)*d) + (Csc[c + d*x]^2*Sqrt[a + b*Sin[c + d*
x]^4])/(a*d) - (Csc[c + d*x]^4*Sqrt[a + b*Sin[c + d*x]^4])/(4*a*d)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

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

Rule 1807

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

Rule 3229

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff^(n/2)*x^(n/2))^p
)/(1 - ff*x)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] &
& IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {\cot ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2}{x^3 \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}-\frac {\operatorname {Subst}\left (\int \frac {4 a-(2 a-b) x}{x^2 \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{4 a d}\\ &=\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}+\frac {(2 a-b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{4 a d}\\ &=\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}+\frac {(2 a-b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^4(c+d x)\right )}{8 a d}\\ &=\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}+\frac {(2 a-b) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^4(c+d x)}\right )}{4 a b d}\\ &=-\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}+\frac {\csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a d}-\frac {\csc ^4(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 2.90, size = 141, normalized size = 1.31 \[ -\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )-4 a \csc ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}+b \sqrt {a+b \sin ^4(c+d x)} \left (\frac {a \csc ^4(c+d x)}{b}-\frac {\tanh ^{-1}\left (\sqrt {\frac {b \sin ^4(c+d x)}{a}+1}\right )}{\sqrt {\frac {b \sin ^4(c+d x)}{a}+1}}\right )}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/4*(2*a^(3/2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]] - 4*a*Csc[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]^4] +
b*Sqrt[a + b*Sin[c + d*x]^4]*((a*Csc[c + d*x]^4)/b - ArcTanh[Sqrt[1 + (b*Sin[c + d*x]^4)/a]]/Sqrt[1 + (b*Sin[c
+ d*x]^4)/a]))/(a^2*d)

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fricas [A]  time = 0.65, size = 371, normalized size = 3.44 \[ \left [-\frac {{\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a - b\right )} \sqrt {a} \log \left (\frac {8 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {a} + 2 \, a + b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (4 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}}, \frac {{\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} + 2 \, a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (4 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(((2*a - b)*cos(d*x + c)^4 - 2*(2*a - b)*cos(d*x + c)^2 + 2*a - b)*sqrt(a)*log(8*(b*cos(d*x + c)^4 - 2*b
*cos(d*x + c)^2 + 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sqrt(a) + 2*a + b)/(cos(d*x + c)^4 - 2
*cos(d*x + c)^2 + 1)) + 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*(4*a*cos(d*x + c)^2 - 3*a))/(a^2
*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d), 1/4*(((2*a - b)*cos(d*x + c)^4 - 2*(2*a - b)*cos(d*x + c)
^2 + 2*a - b)*sqrt(-a)*arctan(sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sqrt(-a)/a) - sqrt(b*cos(d*x
+ c)^4 - 2*b*cos(d*x + c)^2 + a + b)*(4*a*cos(d*x + c)^2 - 3*a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)
^2 + a^2*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )^{5}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)^5/sqrt(b*sin(d*x + c)^4 + a), x)

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maple [F]  time = 1.88, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

int(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)

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maxima [A]  time = 0.69, size = 166, normalized size = 1.54 \[ -\frac {\frac {2 \, \sqrt {b \sin \left (d x + c\right )^{4} + a} b}{{\left (b \sin \left (d x + c\right )^{4} + a\right )} a - a^{2}} - \frac {2 \, \log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {b \log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {8 \, \sqrt {b \sin \left (d x + c\right )^{4} + a}}{a \sin \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

-1/8*(2*sqrt(b*sin(d*x + c)^4 + a)*b/((b*sin(d*x + c)^4 + a)*a - a^2) - 2*log((sqrt(b*sin(d*x + c)^4 + a) - sq
rt(a))/(sqrt(b*sin(d*x + c)^4 + a) + sqrt(a)))/sqrt(a) + b*log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*
sin(d*x + c)^4 + a) + sqrt(a)))/a^(3/2) - 8*sqrt(b*sin(d*x + c)^4 + a)/(a*sin(d*x + c)^2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^5}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(cot(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(cot(c + d*x)**5/sqrt(a + b*sin(c + d*x)**4), x)

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