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

Optimal. Leaf size=123 \[ \frac{10 a \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{10 a \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 d^4 f}+\frac{2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}} \]

[Out]

(-2*b)/(7*f*(d*Sec[e + f*x])^(7/2)) + (10*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])
/(21*d^4*f) + (2*a*Sin[e + f*x])/(7*d*f*(d*Sec[e + f*x])^(5/2)) + (10*a*Sin[e + f*x])/(21*d^3*f*Sqrt[d*Sec[e +
f*x]])

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Rubi [A]  time = 0.0905019, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3486, 3769, 3771, 2641} \[ \frac{10 a \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{10 a \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{d \sec (e+f x)}}{21 d^4 f}+\frac{2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b)/(7*f*(d*Sec[e + f*x])^(7/2)) + (10*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])
/(21*d^4*f) + (2*a*Sin[e + f*x])/(7*d*f*(d*Sec[e + f*x])^(5/2)) + (10*a*Sin[e + f*x])/(21*d^3*f*Sqrt[d*Sec[e +
f*x]])

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)}{(d \sec (e+f x))^{7/2}} \, dx &=-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}}+a \int \frac{1}{(d \sec (e+f x))^{7/2}} \, dx\\ &=-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}}+\frac{2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}+\frac{(5 a) \int \frac{1}{(d \sec (e+f x))^{3/2}} \, dx}{7 d^2}\\ &=-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}}+\frac{2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}+\frac{10 a \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{(5 a) \int \sqrt{d \sec (e+f x)} \, dx}{21 d^4}\\ &=-\frac{2 b}{7 f (d \sec (e+f x))^{7/2}}+\frac{2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}+\frac{10 a \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}+\frac{\left (5 a \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{2 b}{7 f (d \sec (e+f x))^{7/2}}+\frac{10 a \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 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}+\frac{10 a \sin (e+f x)}{21 d^3 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.348078, size = 94, normalized size = 0.76 \[ \frac{\sqrt{d \sec (e+f x)} \left (26 a \sin (2 (e+f x))+3 a \sin (4 (e+f x))+40 a \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )-12 b \cos (2 (e+f x))-3 b \cos (4 (e+f x))-9 b\right )}{84 d^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*Sec[e + f*x]]*(-9*b - 12*b*Cos[2*(e + f*x)] - 3*b*Cos[4*(e + f*x)] + 40*a*Sqrt[Cos[e + f*x]]*EllipticF
[(e + f*x)/2, 2] + 26*a*Sin[2*(e + f*x)] + 3*a*Sin[4*(e + f*x)]))/(84*d^4*f)

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Maple [C]  time = 0.199, size = 190, normalized size = 1.5 \begin{align*}{\frac{2}{21\,f \left ( \cos \left ( fx+e \right ) \right ) ^{4}} \left ( 5\,i\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}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) a+3\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}a-3\,b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+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}}}{\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 \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))/(d*sec(f*x+e))^(7/2),x)

[Out]

2/21/f*(5*I*cos(f*x+e)*(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)/s
in(f*x+e),I)*a+3*sin(f*x+e)*cos(f*x+e)^3*a-3*b*cos(f*x+e)^4+5*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)*a+5*sin(f*x+e)*cos(f*x+e)*a)/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{b \tan \left (f x + e\right ) + a}{\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))/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)/(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{\sqrt{d \sec \left (f x + e\right )}{\left (b \tan \left (f x + e\right ) + a\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))/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)/(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))/(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{b \tan \left (f x + e\right ) + a}{\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))/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

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