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]

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

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Rubi [A]  time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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 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 ^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*}

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Mathematica [A]  time = 0.42, 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]

-1/3*(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])/(Sq
rt[2]*a^2*f)

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fricas [A]  time = 0.64, size = 90, normalized size = 1.22 \[ -\frac {{\left (2 \, {\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 2 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )} \sin \left (f x + e\right )} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^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)

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maple [A]  time = 1.08, size = 86, normalized size = 1.16 \[ \frac {\left (2 a \left (\cos ^{2}\left (f x +e \right )\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) b -3 a +2 b \right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right )}{3 f \sin \left (f x +e \right )^{3} a^{2}} \]

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*(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)*cos(f
*x+e)/sin(f*x+e)^3/a^2

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maxima [A]  time = 0.78, size = 87, normalized size = 1.18 \[ -\frac {\frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )} - \frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{a^{2} \tan \left (f x + e\right )} + \frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{a \tan \left (f x + e\right )^{3}}}{3 \, f} \]

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]

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

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mupad [B]  time = 18.82, size = 145, normalized size = 1.96 \[ -\frac {2\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}}\,\left (a\,1{}\mathrm {i}-b\,1{}\mathrm {i}-a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}-b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}\right )}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(a + b*tan(e + f*x)^2)^(1/2)),x)

[Out]

-(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)*(a*1i -
b*1i - a*exp(e*2i + f*x*2i)*4i + a*exp(e*4i + f*x*4i)*1i + b*exp(e*2i + f*x*2i)*2i - b*exp(e*4i + f*x*4i)*1i)
)/(3*a^2*f*(exp(e*2i + f*x*2i) - 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**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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