3.586 \(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=143 \[ \frac{2 d^2 \left (7 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 f}+\frac{2 d \left (7 a^2-2 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{3/2}}{21 f}+\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

[Out]

(2*(7*a^2 - 2*b^2)*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f) + (18*a*b*(d*
Sec[e + f*x])^(5/2))/(35*f) + (2*(7*a^2 - 2*b^2)*d*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(21*f) + (2*b*(d*Sec[e
+ f*x])^(5/2)*(a + b*Tan[e + f*x]))/(7*f)

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Rubi [A]  time = 0.160003, antiderivative size = 143, 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, 3768, 3771, 2641} \[ \frac{2 d^2 \left (7 a^2-2 b^2\right ) \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 f}+\frac{2 d \left (7 a^2-2 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{3/2}}{21 f}+\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^2,x]

[Out]

(2*(7*a^2 - 2*b^2)*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f) + (18*a*b*(d*
Sec[e + f*x])^(5/2))/(35*f) + (2*(7*a^2 - 2*b^2)*d*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(21*f) + (2*b*(d*Sec[e
+ f*x])^(5/2)*(a + b*Tan[e + f*x]))/(7*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx &=\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac{2}{7} \int (d \sec (e+f x))^{5/2} \left (\frac{7 a^2}{2}-b^2+\frac{9}{2} a b \tan (e+f x)\right ) \, dx\\ &=\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac{1}{7} \left (7 a^2-2 b^2\right ) \int (d \sec (e+f x))^{5/2} \, dx\\ &=\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac{1}{21} \left (\left (7 a^2-2 b^2\right ) d^2\right ) \int \sqrt{d \sec (e+f x)} \, dx\\ &=\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac{1}{21} \left (\left (7 a^2-2 b^2\right ) d^2 \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx\\ &=\frac{2 \left (7 a^2-2 b^2\right ) d^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 f}+\frac{18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac{2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac{2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\\ \end{align*}

Mathematica [A]  time = 0.771574, size = 127, normalized size = 0.89 \[ \frac{2 d^2 \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2 \left (\frac{5}{2} \left (7 a^2-2 b^2\right ) \sin (2 (e+f x))+5 \left (7 a^2-2 b^2\right ) \cos ^{\frac{5}{2}}(e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+3 b (14 a+5 b \tan (e+f x))\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^2,x]

[Out]

(2*d^2*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2*(5*(7*a^2 - 2*b^2)*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2
, 2] + (5*(7*a^2 - 2*b^2)*Sin[2*(e + f*x)])/2 + 3*b*(14*a + 5*b*Tan[e + f*x])))/(105*f*(a*Cos[e + f*x] + b*Sin
[e + f*x])^2)

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Maple [C]  time = 0.313, size = 382, normalized size = 2.7 \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}{105\,f \left ( \sin \left ( fx+e \right ) \right ) ^{4}\cos \left ( fx+e \right ) } \left ( 35\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}} \left ( \cos \left ( fx+e \right ) \right ) ^{4}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ){a}^{2}-10\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}} \left ( \cos \left ( fx+e \right ) \right ) ^{4}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ){b}^{2}+35\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}} \left ( \cos \left ( fx+e \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ){a}^{2}-10\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}} \left ( \cos \left ( fx+e \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ){b}^{2}+35\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}-10\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+42\,\cos \left ( fx+e \right ) ab+15\,\sin \left ( fx+e \right ){b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^2,x)

[Out]

2/105/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(35*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos
(f*x+e)^4*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^2-10*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1
))^(1/2)*cos(f*x+e)^4*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2+35*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(
cos(f*x+e)+1))^(1/2)*cos(f*x+e)^3*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^2-10*I*(1/(cos(f*x+e)+1))^(1/2)*(
cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^3*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2+35*sin(f*x+e)*cos(f
*x+e)^2*a^2-10*sin(f*x+e)*cos(f*x+e)^2*b^2+42*cos(f*x+e)*a*b+15*sin(f*x+e)*b^2)*(d/cos(f*x+e))^(5/2)/sin(f*x+e
)^4/cos(f*x+e)

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Maxima [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((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{2} + 2 \, a b d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right ) + a^{2} d^{2} \sec \left (f x + e\right )^{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))^(5/2)*(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((b^2*d^2*sec(f*x + e)^2*tan(f*x + e)^2 + 2*a*b*d^2*sec(f*x + e)^2*tan(f*x + e) + a^2*d^2*sec(f*x + e)
^2)*sqrt(d*sec(f*x + e)), 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((d*sec(f*x+e))**(5/2)*(a+b*tan(f*x+e))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}{\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))^(5/2)*(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e) + a)^2, x)