3.127 \(\int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=123 \[ -\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \]

[Out]

-1/15*(15*a^2-20*a*b+8*b^2)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/a^3/f-2/15*(5*a-2*b)*cot(f*x+e)^3*(a+b*tan(f*x
+e)^2)^(1/2)/a^2/f-1/5*cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2)/a/f

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Rubi [A]  time = 0.14, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3663, 462, 453, 264} \[ -\frac {\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 - 20*a*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) - (2*(5*a - 2*b)*Cot[e + f*x]^
3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a*f)

Rule 264

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

Rule 453

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

Rule 462

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

Rule 3663

Int[sin[(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^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\csc ^6(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}+\frac {\operatorname {Subst}\left (\int \frac {2 (5 a-2 b)+5 a x^2}{x^4 \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}+\frac {\left (15 a^2-20 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac {2 (5 a-2 b) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{5 a f}\\ \end {align*}

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Mathematica [A]  time = 1.91, size = 90, normalized size = 0.73 \[ -\frac {\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+4 a (a-b) \csc ^2(e+f x)+8 (a-b)^2\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{15 \sqrt {2} a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Cot[e + f*x]*(8*(a - b)^2 + 4*a*(a - b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4)*Sqrt[(a + b + (a - b)*Co
s[2*(e + f*x)])*Sec[e + f*x]^2])/(Sqrt[2]*a^3*f)

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fricas [A]  time = 1.11, size = 141, normalized size = 1.15 \[ -\frac {{\left (8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (5 \, a^{2} - 9 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(5*a^2 - 9*a*b + 4*b^2)*cos(f*x + e)^3 + (15*a^2 - 20*a*b + 8*
b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x
+ e)^2 + a^3*f)*sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.33, size = 148, normalized size = 1.20 \[ -\frac {\left (8 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-16 \left (\cos ^{4}\left (f x +e \right )\right ) a b +8 \left (\cos ^{4}\left (f x +e \right )\right ) b^{2}-20 a^{2} \left (\cos ^{2}\left (f x +e \right )\right )+36 \left (\cos ^{2}\left (f x +e \right )\right ) a b -16 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )+15 a^{2}-20 a b +8 b^{2}\right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right )}{15 f \sin \left (f x +e \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/f*(8*cos(f*x+e)^4*a^2-16*cos(f*x+e)^4*a*b+8*cos(f*x+e)^4*b^2-20*a^2*cos(f*x+e)^2+36*cos(f*x+e)^2*a*b-16*
b^2*cos(f*x+e)^2+15*a^2-20*a*b+8*b^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*cos(f*x+e)/sin(f*
x+e)^5/a^3

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maxima [A]  time = 0.78, size = 173, normalized size = 1.41 \[ -\frac {\frac {15 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )} - \frac {20 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{a^{2} \tan \left (f x + e\right )} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{2}}{a^{3} \tan \left (f x + e\right )} + \frac {10 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )^{3}} - \frac {4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{a^{2} \tan \left (f x + e\right )^{3}} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*sqrt(b*tan(f*x + e)^2 + a)/(a*tan(f*x + e)) - 20*sqrt(b*tan(f*x + e)^2 + a)*b/(a^2*tan(f*x + e)) + 8
*sqrt(b*tan(f*x + e)^2 + a)*b^2/(a^3*tan(f*x + e)) + 10*sqrt(b*tan(f*x + e)^2 + a)/(a*tan(f*x + e)^3) - 4*sqrt
(b*tan(f*x + e)^2 + a)*b/(a^2*tan(f*x + e)^3) + 3*sqrt(b*tan(f*x + e)^2 + a)/(a*tan(f*x + e)^5))/f

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mupad [B]  time = 22.11, size = 761, normalized size = 6.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a - b)*(32*a*b - 64*a^2 + 32*b^2))/(120*a^3*f*(a*1i - b*1i)) - ((a - b)*(64*a^2 - 96*a*b + 32*b^2))/(120*a
^3*f*(a*1i - b*1i)))*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*(2*exp(e*2i + f
*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + f*x*2i) - 1)^2*(exp(e*2i + f*x*2i) + 1)) + ((a + (b*(exp(e*2i +
f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*((2*(3*a - 3*b))/(3*a*f*(a*1i - b*1i)) + ((3*a - 3*b)*(
96*a - 64*b))/(240*a^2*f*(a*1i - b*1i)) + ((3*a - 3*b)*(256*a + 64*b))/(240*a^2*f*(a*1i - b*1i)))*(2*exp(e*2i
+ f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i + f*x*2i) - 1)^4*(exp(e*2i + f*x*2i) + 1)) + ((((a - b)*(32*a
- 16*b))/(30*a^2*f*(a*1i - b*1i)) + ((a - b)*(32*a + 48*b))/(30*a^2*f*(a*1i - b*1i)))*(a + (b*(exp(e*2i + f*x*
2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/((exp(e*2i
+ f*x*2i) - 1)^3*(exp(e*2i + f*x*2i) + 1)) - ((a - b)^2*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*
x*2i) + 1)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1)*8i)/(15*a^3*f*(exp(e*2i + f*x*2i) - 1)*(ex
p(e*2i + f*x*2i) + 1)) + (8*(2*a - 2*b)*(a + (b*(exp(e*2i + f*x*2i)*1i - 1i)^2)/(exp(e*2i + f*x*2i) + 1)^2)^(1
/2)*(2*exp(e*2i + f*x*2i) + exp(e*4i + f*x*4i) + 1))/(5*a*f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*2i + f*x*2i) + 1
)*(a*1i - b*1i))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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