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

Optimal. Leaf size=132 \[ -\frac{\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f (a+b)}-\frac{2 (5 a+3 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

[Out]

-((15*a^2 + 10*a*b + 3*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^3*f) - (2*(5*a + 3*b)*Cot
[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^2*f) - (Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])
/(5*(a + b)*f)

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Rubi [A]  time = 0.139957, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4132, 462, 453, 264} \[ -\frac{\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^3}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f (a+b)}-\frac{2 (5 a+3 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 + 10*a*b + 3*b^2)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^3*f) - (2*(5*a + 3*b)*Cot
[e + f*x]^3*Sqrt[a + b + b*Tan[e + f*x]^2])/(15*(a + b)^2*f) - (Cot[e + f*x]^5*Sqrt[a + b + b*Tan[e + f*x]^2])
/(5*(a + b)*f)

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{2 (5 a+3 b)+5 (a+b) x^2}{x^4 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{2 (5 a+3 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}+\frac{\left (15 a^2+10 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{\left (15 a^2+10 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^3 f}-\frac{2 (5 a+3 b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.350556, size = 100, normalized size = 0.76 \[ -\frac{\csc ^5(e+f x) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a^2 \cos (4 (e+f x))+8 a^2-2 a (3 a+b) \cos (2 (e+f x))+8 a b+3 b^2\right )}{30 f (a+b)^3 \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(8*a^2 + 8*a*b + 3*b^2 - 2*a*(3*a + b)*Cos[2*(e + f*x)] + a^2*Cos[4*(e + f*x)
])*Csc[e + f*x]^5*Sec[e + f*x])/(30*(a + b)^3*f*Sqrt[a + b*Sec[e + f*x]^2])

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Maple [A]  time = 0.383, size = 101, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab+15\,{a}^{2}+10\,ab+3\,{b}^{2} \right ) \cos \left ( fx+e \right ) }{15\,f \left ( a+b \right ) ^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}}\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81973, size = 409, normalized size = 3.1 \begin{align*} -\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \,{\left (5 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} + 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{6}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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