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

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

[Out]

-2/7*b/f/(d*sec(f*x+e))^(7/2)+2/7*a*sin(f*x+e)/d/f/(d*sec(f*x+e))^(5/2)+10/21*a*sin(f*x+e)/d^3/f/(d*sec(f*x+e)
)^(1/2)+10/21*a*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticF(sin(1/2*e+1/2*f*x),2^(1/2))*cos(f*x+
e)^(1/2)*(d*sec(f*x+e))^(1/2)/d^4/f

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.35, 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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fricas [F]  time = 2.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}{d^{4} \sec \left (f x + e\right )^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.82, size = 190, normalized size = 1.54 \[ \frac {\frac {10 i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) a}{21}+\frac {10 i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a}{21}-\frac {2 b \left (\cos ^{4}\left (f x +e \right )\right )}{7}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a}{7}+\frac {10 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a}{21}}{f \cos \left (f x +e \right )^{4} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}} \]

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*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/
2)*cos(f*x+e)*a+5*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*a-3*b*cos(f*x+e)^4+3*cos(f*x+e)^3*sin(f*x+e)*a+5*cos(f*x+e)*sin(f*x+e)*a)/cos(f*x+e)^4/(d/cos(f*x+e
))^(7/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \]

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]

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

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