3.305 \(\int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=167 \[ -\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f} \]

[Out]

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*(a-b)^(1/2)/f-1/15*(15*a^2-5*a*b-2*b^2)*cot(f*x+e)*(a
+b*tan(f*x+e)^2)^(1/2)/a^2/f+1/15*(5*a-b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/a/f-1/5*cot(f*x+e)^5*(a+b*tan(
f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.23, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3670, 475, 583, 12, 377, 203} \[ -\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-((Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f) - ((15*a^2 - 5*a*b - 2*b^2)*C
ot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^2*f) + ((5*a - b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(15
*a*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \cot ^6(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {\operatorname {Subst}\left (\int \frac {-5 a+b-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {\operatorname {Subst}\left (\int \frac {-15 a^2+5 a b+2 b^2-2 (5 a-b) b x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {\operatorname {Subst}\left (\int -\frac {15 a^2 (a-b)}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {a-b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac {(5 a-b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}\\ \end {align*}

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Mathematica [C]  time = 14.39, size = 339, normalized size = 2.03 \[ -\frac {\cos ^4(e+f x) \cot ^5(e+f x) \left (\frac {b \tan ^2(e+f x)}{a}+1\right ) \left (8 (a-b) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^3 \, _3F_2\left (2,2,2;1,\frac {3}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )+8 \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \left (-2 a^2+a b \left (3 \tan ^2(e+f x)+2\right )-3 b^2 \tan ^2(e+f x)\right ) \, _2F_1\left (2,2;\frac {3}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )+\frac {a^2 \sec ^4(e+f x) \left (3 a^2-4 a b \tan ^2(e+f x)+8 b^2 \tan ^4(e+f x)\right ) \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}} \sin ^{-1}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right )+\sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}\right )}{\sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}\right )}{15 a^3 f \sqrt {a+b \tan ^2(e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^6*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-1/15*(Cos[e + f*x]^4*Cot[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*(8*(a - b)*HypergeometricPFQ[{2, 2, 2}, {1, 3/
2}, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^3 + 8*Hypergeometric2F1[2, 2, 3/2, ((a -
b)*Sin[e + f*x]^2)/a]*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^2*(-2*a^2 - 3*b^2*Tan[e + f*x]^2 + a*b*(2 + 3*Tan
[e + f*x]^2)) + (a^2*Sec[e + f*x]^4*(3*a^2 - 4*a*b*Tan[e + f*x]^2 + 8*b^2*Tan[e + f*x]^4)*(ArcSin[Sqrt[((a - b
)*Sin[e + f*x]^2)/a]]*Sqrt[((a - b)*Sin[e + f*x]^2)/a] + Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]))/Sqr
t[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]))/(a^3*f*Sqrt[a + b*Tan[e + f*x]^2])

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fricas [A]  time = 0.54, size = 375, normalized size = 2.25 \[ \left [\frac {15 \, a^{2} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} - 4 \, {\left ({\left (15 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, a^{2} f \tan \left (f x + e\right )^{5}}, -\frac {15 \, \sqrt {a - b} a^{2} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{2} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, a^{2} f \tan \left (f x + e\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/60*(15*a^2*sqrt(-a + b)*log(-((a^2 - 8*a*b + 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2
- 4*((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*
tan(f*x + e)^2 + 1))*tan(f*x + e)^5 - 4*((15*a^2 - 5*a*b - 2*b^2)*tan(f*x + e)^4 - (5*a^2 - a*b)*tan(f*x + e)^
2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a^2*f*tan(f*x + e)^5), -1/30*(15*sqrt(a - b)*a^2*arctan(-2*sqrt(b*tan(
f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a))*tan(f*x + e)^5 + 2*((15*a^2 - 5*a*b -
2*b^2)*tan(f*x + e)^4 - (5*a^2 - a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a^2*f*tan(f*x + e)
^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^6, x)

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maple [C]  time = 1.36, size = 6894, normalized size = 41.28 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (e+f\,x\right )}^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**6*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*cot(e + f*x)**6, x)

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