3.5.92 \(\int \cot ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [492]

Optimal. Leaf size=110 \[ \frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {(2 a-b) \sqrt {a+b \sin ^2(e+f x)}}{2 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 a f} \]

[Out]

-1/2*csc(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2)/a/f+1/2*(2*a-b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)
-1/2*(2*a-b)*(a+b*sin(f*x+e)^2)^(1/2)/a/f

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Rubi [A]
time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 52, 65, 214} \begin {gather*} -\frac {(2 a-b) \sqrt {a+b \sin ^2(e+f x)}}{2 a f}+\frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{2 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

((2*a - b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*Sqrt[a]*f) - ((2*a - b)*Sqrt[a + b*Sin[e + f*x]^2])
/(2*a*f) - (Csc[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2))/(2*a*f)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^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]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 77, normalized size = 0.70 \begin {gather*} \frac {(2 a-b) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )-\sqrt {a} \left (2+\csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{2 \sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

((2*a - b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] - Sqrt[a]*(2 + Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^
2])/(2*Sqrt[a]*f)

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Maple [A]
time = 9.70, size = 122, normalized size = 1.11

method	result	size
default	\(\frac {-\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 \sin \left (f x +e \right )^{2}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2 \sqrt {a}}+\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-(a+b*sin(f*x+e)^2)^(1/2)-1/2/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)-1/2/a^(1/2)*b*ln((2*a+2*a^(1/2)*(a+b*sin(
f*x+e)^2)^(1/2))/sin(f*x+e))+a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)))/f

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Maxima [A]
time = 0.33, size = 119, normalized size = 1.08 \begin {gather*} \frac {2 \, \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \frac {b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a} - \frac {{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*(2*sqrt(a)*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/sqrt(a) -
2*sqrt(b*sin(f*x + e)^2 + a) + sqrt(b*sin(f*x + e)^2 + a)*b/a - (b*sin(f*x + e)^2 + a)^(3/2)/(a*sin(f*x + e)^
2))/f

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Fricas [A]
time = 0.96, size = 239, normalized size = 2.17 \begin {gather*} \left [-\frac {{\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left (2 \, a \cos \left (f x + e\right )^{2} - 3 \, a\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}, -\frac {{\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) + {\left (2 \, a \cos \left (f x + e\right )^{2} - 3 \, a\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cot ^{3}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*cot(e + f*x)**3, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{4096,[8,8]%%%}+%%%{%%%{16384,[1]%%%},[8,7]%%%}+%%%{%%%{2
4576,[2]%%%

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cot}\left (e+f\,x\right )}^3\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^3*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cot(e + f*x)^3*(a + b*sin(e + f*x)^2)^(1/2), x)

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