3.126 \(\int \frac{\csc ^4(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{(3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac{\cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.0900209, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3663, 453, 264} \[ -\frac{(3 a-2 b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a^2 f}-\frac{\cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

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]

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

Mathematica [A]  time = 0.413838, size = 68, normalized size = 0.92 \[ -\frac{\cot (e+f x) \left (a \csc ^2(e+f x)+2 a-2 b\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{3 \sqrt{2} a^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.194, size = 86, normalized size = 1.2 \begin{align*}{\frac{ \left ( 2\,a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}b-3\,a+2\,b \right ) \cos \left ( fx+e \right ) }{3\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \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)^4/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

1/3/f/a^2*(2*a*cos(f*x+e)^2-2*cos(f*x+e)^2*b-3*a+2*b)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*c
os(f*x+e)/sin(f*x+e)^3

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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)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.12206, size = 211, normalized size = 2.85 \begin{align*} -\frac{{\left (2 \,{\left (a - b\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a - 2 \, b\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}}}}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} 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)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (e + f x \right )}}{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(csc(e + f*x)**4/sqrt(a + b*tan(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{4}}{\sqrt{b \tan \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)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/sqrt(b*tan(f*x + e)^2 + a), x)