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

Optimal. Leaf size=94 \[ \frac{6 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0704951, antiderivative size = 94, 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, 3769, 3771, 2639} \[ \frac{6 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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)}{(d \sec (e+f x))^{5/2}} \, dx &=-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}}+a \int \frac{1}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac{2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}+\frac{(3 a) \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac{2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}+\frac{(3 a) \int \sqrt{\cos (e+f x)} \, dx}{5 d^2 \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac{6 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.605078, size = 74, normalized size = 0.79 \[ \frac{2 \sqrt{d \sec (e+f x)} \left (\cos ^2(e+f x) (a \sin (e+f x)-b \cos (e+f x))+3 a \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{5 d^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.204, size = 345, normalized size = 3.7 \begin{align*} -{\frac{2}{5\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( 3\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) a\sin \left ( fx+e \right ) -3\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\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 ) a\sin \left ( fx+e \right ) +3\,i\sin \left ( fx+e \right ){\it EllipticE} \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-3\,i\sin \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{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}a+\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}b+ \left ( \cos \left ( fx+e \right ) \right ) ^{4}a+2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a-3\,a\cos \left ( fx+e \right ) \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((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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