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

Optimal. Leaf size=66 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3663, 277, 217, 206} \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2*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

Rule 206

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

Rule 217

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 277

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 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 \csc ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {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 {b \operatorname {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 {b \operatorname {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} \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [C]  time = 2.18, size = 156, normalized size = 2.36 \[ -\frac {\tan (e+f x) \left (\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)-\sqrt {2} b \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )\right )}{\sqrt {2} f \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 331, normalized size = 5.02 \[ \left [\frac {\sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{4 \, f \sin \left (f x + e\right )}, -\frac {\sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, f \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(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)/c
os(f*x + e)^4)*sin(f*x + e) - 4*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(f*sin(f*x + e
)), -1/2*(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*sqrt(((a - b)*cos
(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(f*sin(f*x + e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(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)^2, x)

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maple [C]  time = 4.52, size = 1215, normalized size = 18.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*cos(f*x+e)*(-2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^
(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)
*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticF((-1+cos(f*x+e))*
((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8
*a*b+8*b^2)/a^2)^(1/2))*b*sin(f*x+e)*cos(f*x+e)+2*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(
1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*
b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1
/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/
((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2))*b*sin(f*x+e)*cos(f*x+e)-sin(f*x+e)*EllipticF((-1+cos(f*x+e))*((2*I*
(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*(a-b)^(1/2)*b^(3/2)-4*I*(a-b)^(1/2)*b^(1/2)*a+a^2-8*a*b+8
*b^2)/a^2)^(1/2))*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b
)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x+e)+b*cos(f*x+e
)-b)/(1+cos(f*x+e))/a)^(1/2)*b+2*2^(1/2)*((I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)+a*cos(f*x+e)
-b*cos(f*x+e)+b)/(1+cos(f*x+e))/a)^(1/2)*(-2*(I*cos(f*x+e)*(a-b)^(1/2)*b^(1/2)-I*(a-b)^(1/2)*b^(1/2)-a*cos(f*x
+e)+b*cos(f*x+e)-b)/(1+cos(f*x+e))/a)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/
2)/sin(f*x+e),1/(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)*a,(-(2*I*(a-b)^(1/2)*b^(1/2)-a+2*b)/a)^(1/2)/((2*I*(a-b)^(1/2)
*b^(1/2)+a-2*b)/a)^(1/2))*b*sin(f*x+e)+cos(f*x+e)^2*((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*a-cos(f*x+e)^2*(
(2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b+((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)*b)/sin(f*x+e)/(a*cos(f*x+
e)^2-cos(f*x+e)^2*b+b)/((2*I*(a-b)^(1/2)*b^(1/2)+a-2*b)/a)^(1/2)

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maxima [A]  time = 0.63, size = 47, normalized size = 0.71 \[ \frac {\sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) - \frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\tan \left (f x + e\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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