3.588 \(\int \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=103 \[ \frac{2 \left (3 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 f}+\frac{10 a b \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f} \]

[Out]

(10*a*b*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*(3*a^2 - 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*S
ec[e + f*x]])/(3*f) + (2*b*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x]))/(3*f)

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Rubi [A]  time = 0.123328, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3508, 3486, 3771, 2641} \[ \frac{2 \left (3 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 f}+\frac{10 a b \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2,x]

[Out]

(10*a*b*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*(3*a^2 - 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*S
ec[e + f*x]])/(3*f) + (2*b*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x]))/(3*f)

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx &=\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f}+\frac{2}{3} \int \sqrt{d \sec (e+f x)} \left (\frac{3 a^2}{2}-b^2+\frac{5}{2} a b \tan (e+f x)\right ) \, dx\\ &=\frac{10 a b \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f}+\frac{1}{3} \left (3 a^2-2 b^2\right ) \int \sqrt{d \sec (e+f x)} \, dx\\ &=\frac{10 a b \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f}+\frac{1}{3} \left (\left (3 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx\\ &=\frac{10 a b \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 \left (3 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.617032, size = 87, normalized size = 0.84 \[ \frac{2 \sec ^2(e+f x) \sqrt{d \sec (e+f x)} \left (\left (3 a^2-2 b^2\right ) \cos ^{\frac{5}{2}}(e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+b \cos (e+f x) (6 a \cos (e+f x)+b \sin (e+f x))\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2,x]

[Out]

(2*Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*((3*a^2 - 2*b^2)*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + b*Cos[e
+ f*x]*(6*a*Cos[e + f*x] + b*Sin[e + f*x])))/(3*f)

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Maple [C]  time = 0.282, size = 339, normalized size = 3.3 \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{3\,f\cos \left ( fx+e \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{4}}\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}} \left ( 3\,i \left ( \cos \left ( fx+e \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}{a}^{2}-2\,i \left ( \cos \left ( fx+e \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}{b}^{2}+3\,i\cos \left ( fx+e \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}{a}^{2}-2\,i\cos \left ( fx+e \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}{b}^{2}+6\,\cos \left ( fx+e \right ) ab+\sin \left ( fx+e \right ){b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/f*(d/cos(f*x+e))^(1/2)*(cos(f*x+e)-1)^2*(3*I*cos(f*x+e)^2*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(cos(f*
x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*a^2-2*I*cos(f*x+e)^2*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e)
,I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*b^2+3*I*cos(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/s
in(f*x+e),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*a^2-2*I*cos(f*x+e)*EllipticF(I*(cos(f*
x+e)-1)/sin(f*x+e),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*b^2+6*cos(f*x+e)*a*b+sin(f*x+
e)*b^2)*(cos(f*x+e)+1)^2/cos(f*x+e)/sin(f*x+e)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \sqrt{d \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*sqrt(d*sec(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)^2, x)