3.318 \(\int \cot ^6(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=165 \[ -\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f} \]

[Out]

-(a-b)^(3/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f-1/15*(15*a^2-20*a*b+3*b^2)*cot(f*x+e)*(
a+b*tan(f*x+e)^2)^(1/2)/a/f+1/15*(5*a-6*b)*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2)/f-1/5*a*cot(f*x+e)^5*(a+b*tan
(f*x+e)^2)^(1/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 165, 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, 474, 583, 12, 377, 203} \[ -\frac {\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-(((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f) - ((15*a^2 - 20*a*b + 3*b^2
)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a*f) + ((5*a - 6*b)*Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/
(15*f) - (a*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 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(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 - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[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) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}+\frac {\operatorname {Subst}\left (\int \frac {-a (5 a-6 b)-(4 a-5 b) 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-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {\operatorname {Subst}\left (\int \frac {-a \left (15 a^2-20 a b+3 b^2\right )-2 a (5 a-6 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^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \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)^2}{\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^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {(a-b)^2 \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^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}-\frac {(a-b)^2 \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 {(a-b)^{3/2} \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^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a f}+\frac {(5 a-6 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 f}-\frac {a \cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 9.50, size = 140, normalized size = 0.85 \[ -\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)} \left (a \cot ^2(e+f x)+b\right )^2 \left (2 (a-b) ((a-b) \cos (2 (e+f x))+a+b) \, _2F_1\left (2,2;\frac {1}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )+a \left (3 a \cot ^2(e+f x)-2 b\right ) \, _2F_1\left (1,1;-\frac {1}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right )\right )}{15 a^3 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^6*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-1/15*(Cos[e + f*x]*(b + a*Cot[e + f*x]^2)^2*(a*(-2*b + 3*a*Cot[e + f*x]^2)*Hypergeometric2F1[1, 1, -1/2, ((a
- b)*Sin[e + f*x]^2)/a] + 2*(a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])*Hypergeometric2F1[2, 2, 1/2, ((a - b)*S
in[e + f*x]^2)/a])*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(a^3*f)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 385, normalized size = 2.33 \[ \left [-\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{60 \, a f \tan \left (f x + e\right )^{5}}, -\frac {15 \, {\left (a^{2} - a b\right )} \sqrt {a - b} \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \, {\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - {\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{30 \, a f \tan \left (f x + e\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/60*(15*(a^2 - a*b)*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 - 20*a*b + 3*b^2)*tan(f*x + e)^4 - (5*a^2 - 6*a*b)*t
an(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*f*tan(f*x + e)^5), -1/30*(15*(a^2 - a*b)*sqrt(a - b)*arc
tan(-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 - 20*a*b + 3*b^2)*tan(f*x + e)^4 - (5*a^2 - 6*a*b)*tan(f*x + e)^2 + 3*a^2)*sqrt(b*tan(f*x + e)^2 + a)
)/(a*f*tan(f*x + e)^5)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)

________________________________________________________________________________________

maple [C]  time = 1.41, size = 10026, normalized size = 60.76 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^6, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (e+f\,x\right )}^6\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(cot(e + f*x)^6*(a + b*tan(e + f*x)^2)^(3/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**6*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________