3.6.92 \(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx\) [592]

Optimal. Leaf size=184 \[ -\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{21 d^4 f}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}} \]

[Out]

-6/35*a*b/f/(d*sec(f*x+e))^(7/2)+2/35*(5*a^2+2*b^2)*sin(f*x+e)/d/f/(d*sec(f*x+e))^(5/2)+2/21*(5*a^2+2*b^2)*sin
(f*x+e)/d^3/f/(d*sec(f*x+e))^(1/2)+2/21*(5*a^2+2*b^2)*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Elliptic
F(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-2/5*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+
e))^(7/2)

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Rubi [A]
time = 0.15, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3589, 3567, 3854, 3856, 2720} \begin {gather*} \frac {2 \left (5 a^2+2 b^2\right ) \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 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*b)/(35*f*(d*Sec[e + f*x])^(7/2)) + (2*(5*a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[
d*Sec[e + f*x]])/(21*d^4*f) + (2*(5*a^2 + 2*b^2)*Sin[e + f*x])/(35*d*f*(d*Sec[e + f*x])^(5/2)) + (2*(5*a^2 + 2
*b^2)*Sin[e + f*x])/(21*d^3*f*Sqrt[d*Sec[e + f*x]]) - (2*b*(a + b*Tan[e + f*x]))/(5*f*(d*Sec[e + f*x])^(7/2))

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 3589

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

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*} \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{7/2}} \, dx &=-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}-\frac {2}{5} \int \frac {-\frac {5 a^2}{2}-b^2-\frac {3}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{7/2}} \, dx\\ &=-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}-\frac {1}{5} \left (-5 a^2-2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{7/2}} \, dx\\ &=-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac {\left (5 a^2+2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{3/2}} \, dx}{7 d^2}\\ &=-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac {\left (5 a^2+2 b^2\right ) \int \sqrt {d \sec (e+f x)} \, dx}{21 d^4}\\ &=-\frac {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}+\frac {\left (\left (5 a^2+2 b^2\right ) \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 {6 a b}{35 f (d \sec (e+f x))^{7/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \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 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{35 d f (d \sec (e+f x))^{5/2}}+\frac {2 \left (5 a^2+2 b^2\right ) \sin (e+f x)}{21 d^3 f \sqrt {d \sec (e+f x)}}-\frac {2 b (a+b \tan (e+f x))}{5 f (d \sec (e+f x))^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 2.46, size = 127, normalized size = 0.69 \begin {gather*} \frac {-18 a b \cos (e+f x)-6 a b \cos (3 (e+f x))+\frac {4 \left (5 a^2+2 b^2\right ) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}+23 a^2 \sin (e+f x)+5 b^2 \sin (e+f x)+3 a^2 \sin (3 (e+f x))-3 b^2 \sin (3 (e+f x))}{42 d^3 f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-18*a*b*Cos[e + f*x] - 6*a*b*Cos[3*(e + f*x)] + (4*(5*a^2 + 2*b^2)*EllipticF[(e + f*x)/2, 2])/Sqrt[Cos[e + f*
x]] + 23*a^2*Sin[e + f*x] + 5*b^2*Sin[e + f*x] + 3*a^2*Sin[3*(e + f*x)] - 3*b^2*Sin[3*(e + f*x)])/(42*d^3*f*Sq
rt[d*Sec[e + f*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.54, size = 359, normalized size = 1.95

method	result	size
default	\(\frac {\frac {10 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a^{2}}{21}+\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) b^{2}}{21}+\frac {10 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}}{21}+\frac {4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}}{21}-\frac {4 \left (\cos ^{4}\left (f x +e \right )\right ) a b}{7}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{2}}{7}-\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b^{2}}{7}+\frac {10 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2}}{21}+\frac {4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b^{2}}{21}}{f \cos \left (f x +e \right )^{4} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}\)	\(359\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21/f*(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)*EllipticF(I*(cos(f*x+e)-1)/s
in(f*x+e),I)*a^2+2*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*EllipticF(I*(cos(f*
x+e)-1)/sin(f*x+e),I)*b^2+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^2+2*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)*b^2-6*cos(f*x+e)^4*a*b+3*cos(f*x+e)^3*sin(f*x+e)*a^2-3*cos(f*x+e)^3*sin(f*x+e)*b^2+5*cos(f*x
+e)*sin(f*x+e)*a^2+2*cos(f*x+e)*sin(f*x+e)*b^2)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 163, normalized size = 0.89 \begin {gather*} \frac {\sqrt {2} {\left (-5 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (5 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (6 \, a b \cos \left (f x + e\right )^{4} - {\left (3 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{2} + 2 \, b^{2}\right )} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(sqrt(2)*(-5*I*a^2 - 2*I*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2)
*(5*I*a^2 + 2*I*b^2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(6*a*b*cos(f*x + e)
^4 - (3*(a^2 - b^2)*cos(f*x + e)^3 + (5*a^2 + 2*b^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^4*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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