\(\int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [102]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3744, 462, 283, 223, 212} \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 283

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

Rule 462

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 3744

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 4.58 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.04 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=-\frac {\left (\left (6 a^2+11 a b+3 b^2+4 \left (a^2-3 a b-b^2\right ) \cos (2 (e+f x))+\left (-2 a^2+a b+b^2\right ) \cos (4 (e+f x))\right ) \csc ^4(e+f x)-12 \sqrt {2} a b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \tan (e+f x)}{12 \sqrt {2} a f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]

[In]

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

[Out]

-1/12*(((6*a^2 + 11*a*b + 3*b^2 + 4*(a^2 - 3*a*b - b^2)*Cos[2*(e + f*x)] + (-2*a^2 + a*b + b^2)*Cos[4*(e + f*x
)])*Csc[e + f*x]^4 - 12*Sqrt[2]*a*b*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcS
in[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Tan[e + f*x])/(Sqrt[2]*a*f*Sqrt[(
a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(339\) vs. \(2(88)=176\).

Time = 3.98 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.40

method	result	size
default	\(-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}\, \left (3 a \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right ) \cot \left (f x +e \right )^{2}+\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cot \left (f x +e \right )-3 a \sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {b}}\right ) \cot \left (f x +e \right ) \csc \left (f x +e \right )-2 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \cot \left (f x +e \right )^{3}+3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \cot \left (f x +e \right ) \csc \left (f x +e \right )^{2}\right )}{3 f a \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(340\)

[In]

int(csc(f*x+e)^4*(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f/a*(a+b*tan(f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*(3*a*b^(1/2)*arct
anh(1/b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))*cot(f*x+e)^2+(
(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b*cot(f*x+e)-3*a*b^(1/2)*arctanh(1/b^(1/2)*((a*cos(f
*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))*cot(f*x+e)*csc(f*x+e)-2*((a*cos(f*x+e
)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a*cot(f*x+e)^3+3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+
1)^2)^(1/2)*a*cot(f*x+e)*csc(f*x+e)^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (88) = 176\).

Time = 0.45 (sec) , antiderivative size = 435, normalized size of antiderivative = 4.35 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )} \sin \left (f x + e\right )}, -\frac {3 \, {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left (a f \cos \left (f x + e\right )^{2} - a f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[1/12*(3*(a*cos(f*x + e)^2 - a)*sqrt(b)*log(((a^2 - 8*a*b + 8*b^2)*cos(f*x + e)^4 + 8*(a*b - 2*b^2)*cos(f*x +
e)^2 + 4*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^
2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((2*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*
sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a*f*cos(f*x + e)^2 - a*f)*sin(f*x + e)), -1/6*(3*(a*cos(f
*x + e)^2 - a)*sqrt(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)/(((a*b - b^2)*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) + 2*((2*a + b)*c
os(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a*f*cos(f*x + e)^
2 - a*f)*sin(f*x + e))]

Sympy [F]

\[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{4}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {3 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) - \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a}}{\tan \left (f x + e\right )} - \frac {{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \tan \left (f x + e\right )^{3}}}{3 \, f} \]

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

Giac [F]

\[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int { \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^4} \,d x \]

[In]

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

[Out]

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