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

Optimal. Leaf size=78 \[ -\frac{\cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 f (a+b)}-\frac{(3 a+b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 f (a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.0970371, antiderivative size = 78, 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 = {4132, 453, 264} \[ -\frac{\cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 f (a+b)}-\frac{(3 a+b) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{3 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.209077, size = 74, normalized size = 0.95 \[ \frac{\csc ^3(e+f x) \sec (e+f x) (a \cos (2 (e+f x))-2 a-b) (a \cos (2 (e+f x))+a+2 b)}{6 f (a+b)^2 \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a - b + a*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*(e + f*x)])*Csc[e + f*x]^3*Sec[e + f*x])/(6*(a + b)^2*f*Sq
rt[a + b*Sec[e + f*x]^2])

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Maple [A]  time = 0.343, size = 66, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2\,a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,a-b \right ) \cos \left ( fx+e \right ) }{3\,f \left ( a+b \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}}\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)^4/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

1/3/f/(a+b)^2*(2*a*cos(f*x+e)^2-3*a-b)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)*cos(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*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.729017, size = 235, normalized size = 3.01 \begin{align*} -\frac{{\left (2 \, a \cos \left (f x + e\right )^{3} -{\left (3 \, a + b\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}}}}{3 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} -{\left (a^{2} + 2 \, a b + b^{2}\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)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/3*(2*a*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^2 + 2*a*b
+ b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*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)**4/(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 )^{4}}{\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)^4/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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