3.592 \(\int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=184 \[ \frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{2 \left (5 a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 d^4 f}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}} \]

[Out]

(-6*a*b)/(35*f*(d*Sec[e + f*x])^(7/2)) + (2*(5*a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[
d*Sec[e + f*x]])/(21*d^4*f) + (2*(5*a^2 + 2*b^2)*Sin[e + f*x])/(35*d*f*(d*Sec[e + f*x])^(5/2)) + (2*(5*a^2 + 2
*b^2)*Sin[e + f*x])/(21*d^3*f*Sqrt[d*Sec[e + f*x]]) - (2*b*(a + b*Tan[e + f*x]))/(5*f*(d*Sec[e + f*x])^(7/2))

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Rubi [A]  time = 0.191904, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{2 \left (5 a^2+2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 d^4 f}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(7/2),x]

[Out]

(-6*a*b)/(35*f*(d*Sec[e + f*x])^(7/2)) + (2*(5*a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[
d*Sec[e + f*x]])/(21*d^4*f) + (2*(5*a^2 + 2*b^2)*Sin[e + f*x])/(35*d*f*(d*Sec[e + f*x])^(5/2)) + (2*(5*a^2 + 2
*b^2)*Sin[e + f*x])/(21*d^3*f*Sqrt[d*Sec[e + f*x]]) - (2*b*(a + b*Tan[e + f*x]))/(5*f*(d*Sec[e + f*x])^(7/2))

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))^{7/2}} \, dx &=-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}-\frac{2}{5} \int \frac{-\frac{5 a^2}{2}-b^2-\frac{3}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx\\ &=-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}-\frac{1}{5} \left (-5 a^2-2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{7/2}} \, dx\\ &=-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac{\left (5 a^2+2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{3/2}} \, dx}{7 d^2}\\ &=-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac{\left (5 a^2+2 b^2\right ) \int \sqrt{d \sec (e+f x)} \, dx}{21 d^4}\\ &=-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac{\left (\left (5 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}{21 d^4}\\ &=-\frac{6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac{2 \left (5 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)}}{21 d^4 f}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac{2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}-\frac{2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 2.22214, size = 127, normalized size = 0.69 \[ \frac{\frac{4 \left (5 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\sqrt{\cos (e+f x)}}+23 a^2 \sin (e+f x)+3 a^2 \sin (3 (e+f x))-18 a b \cos (e+f x)-6 a b \cos (3 (e+f x))+5 b^2 \sin (e+f x)-3 b^2 \sin (3 (e+f x))}{42 d^3 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(7/2),x]

[Out]

(-18*a*b*Cos[e + f*x] - 6*a*b*Cos[3*(e + f*x)] + (4*(5*a^2 + 2*b^2)*EllipticF[(e + f*x)/2, 2])/Sqrt[Cos[e + f*
x]] + 23*a^2*Sin[e + f*x] + 5*b^2*Sin[e + f*x] + 3*a^2*Sin[3*(e + f*x)] - 3*b^2*Sin[3*(e + f*x)])/(42*d^3*f*Sq
rt[d*Sec[e + f*x]])

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Maple [C]  time = 0.281, size = 359, normalized size = 2. \begin{align*}{\frac{2}{21\,f \left ( \cos \left ( fx+e \right ) \right ) ^{4}} \left ( 5\,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}+5\,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}-6\,ab \left ( \cos \left ( fx+e \right ) \right ) ^{4}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ){a}^{2}-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sin \left ( fx+e \right ){b}^{2}+5\,\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){a}^{2}+2\,\cos \left ( fx+e \right ) \sin \left ( fx+e \right ){b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{7}{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))^(7/2),x)

[Out]

2/21/f*(5*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)/
(cos(f*x+e)+1))^(1/2)*b^2+5*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)/(c
os(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^2-6*a*b*cos(f*x+e)^4+3*cos(f*x+e)^3*sin(f*x+e)*a^2-3*cos(f*x+e)^3*sin(f*x+e)*b^2+5*cos(f*x
+e)*sin(f*x+e)*a^2+2*cos(f*x+e)*sin(f*x+e)*b^2)/cos(f*x+e)^4/(d/cos(f*x+e))^(7/2)

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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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(7/2), 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^{4} \sec \left (f x + e\right )^{4}}, 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))^(7/2),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^4*sec(f*x + e)^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(7/2),x)

[Out]

Timed out

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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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(7/2), x)