3.590 \(\int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}+\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.162471, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3508, 3486, 3769, 3771, 2641} \[ \frac{2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}+\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(3/2),x]

[Out]

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

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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 \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{3/2}} \, dx &=-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}-2 \int \frac{-\frac{a^2}{2}-b^2+\frac{1}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}-\left (-a^2-2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{3/2}} \, dx\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}+\frac{\left (a^2+2 b^2\right ) \int \sqrt{d \sec (e+f x)} \, dx}{3 d^2}\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}+\frac{\left (\left (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}{3 d^2}\\ &=\frac{2 a b}{3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (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 d^2 f}+\frac{2 \left (a^2+2 b^2\right ) \sin (e+f x)}{3 d f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.519407, size = 101, normalized size = 0.73 \[ \frac{\sec ^2(e+f x) \left (2 \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+a^2 \sin (2 (e+f x))-2 a b \cos (2 (e+f x))-2 a b-b^2 \sin (2 (e+f x))\right )}{3 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(3/2),x]

[Out]

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

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Maple [C]  time = 0.277, size = 320, normalized size = 2.3 \begin{align*}{\frac{2}{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{a}^{2}+2\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{b}^{2}+i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{a}^{2}+2\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{b}^{2}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab+\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){a}^{2}-\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(3/2),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))/(d^2*sec(f*x + e)^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}{\left (d \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(3/2), x)