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]

-((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)

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Rubi [A]  time = 0.175569, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, 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 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]

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 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 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 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]

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*}

Mathematica [A]  time = 2.99334, 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]

-(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*Sq
rt[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]))/(4*a^2*d)

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Maple [F]  time = 0.736, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cot \left ( dx+c \right ) \right ) ^{5}{\frac{1}{\sqrt{a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.77907, size = 913, normalized size = 8.45 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\left (c + d x \right )}}{\sqrt{a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{5}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

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)