\(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx\) [585]

Optimal result

Integrand size = 23, antiderivative size = 123 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=-\frac {2 b}{7 f (d \sec (e+f x))^{7/2}}+\frac {10 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f}+\frac {2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}+\frac {10 a \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}} \]

[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*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+
e)^(1/2)*(d*sec(f*x+e))^(1/2)/d^4/f

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3567, 3854, 3856, 2720} \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\frac {10 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f}+\frac {10 a \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{7 d f (d \sec (e+f x))^{5/2}}-\frac {2 b}{7 f (d \sec (e+f x))^{7/2}} \]

[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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3567

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 3854

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 3856

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*} \text {integral}& = -\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)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),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] (verified)

Time = 1.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\frac {\sqrt {d \sec (e+f x)} \left (-9 b-12 b \cos (2 (e+f x))-3 b \cos (4 (e+f x))+40 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+26 a \sin (2 (e+f x))+3 a \sin (4 (e+f x))\right )}{84 d^4 f} \]

[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)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {2 a \left (5 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+5 i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-5 \sin \left (f x +e \right )\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}-\frac {2 b}{7 f \left (d \sec \left (f x +e \right )\right )^{\frac {7}{2}}}\)	\(178\)
parts	\(-\frac {2 a \left (5 i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+5 i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-5 \sin \left (f x +e \right )\right )}{21 f \sqrt {d \sec \left (f x +e \right )}\, d^{3}}-\frac {2 b}{7 f \left (d \sec \left (f x +e \right )\right )^{\frac {7}{2}}}\)	\(178\)

[In]

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\frac {-5 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (3 \, b \cos \left (f x + e\right )^{4} - {\left (3 \, a \cos \left (f x + e\right )^{3} + 5 \, a \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{21 \, d^{4} f} \]

[In]

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

[Out]

1/21*(-5*I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*I*sqrt(2)*a*sqrt(d)
*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(3*b*cos(f*x + e)^4 - (3*a*cos(f*x + e)^3 + 5*a
*cos(f*x + e))*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^4*f)

Sympy [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]

[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)