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

Optimal. Leaf size=92 \[ \frac{2 a d^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 a d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}+\frac{2 b (d \sec (e+f x))^{5/2}}{5 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0691213, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3486, 3768, 3771, 2641} \[ \frac{2 a d^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{3 f}+\frac{2 a d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}+\frac{2 b (d \sec (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.408076, size = 58, normalized size = 0.63 \[ \frac{(d \sec (e+f x))^{5/2} \left (5 a \sin (2 (e+f x))+10 a \cos ^{\frac{5}{2}}(e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+6 b\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sec[e + f*x])^(5/2)*(6*b + 10*a*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + 5*a*Sin[2*(e + f*x)]))/(15*
f)

________________________________________________________________________________________

Maple [C]  time = 0.214, size = 195, normalized size = 2.1 \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}{15\,f \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \left ( 5\,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+5\,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 ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) a+5\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) a+3\,b \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)),x)

[Out]

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

________________________________________________________________________________________

Maxima [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 )}\,{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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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)),x)

[Out]

Timed out

________________________________________________________________________________________

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 )}\,{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)),x, algorithm="giac")

[Out]

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