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

Optimal. Leaf size=143 \[ \frac {2 d^2 \left (7 a^2-2 b^2\right ) \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 d \left (7 a^2-2 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{3/2}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

[Out]

18/35*a*b*(d*sec(f*x+e))^(5/2)/f+2/21*(7*a^2-2*b^2)*d*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/21*(7*a^2-2*b^2)*d^2
*(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)/f+2/7*b*(d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))/f

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Rubi [A]  time = 0.16, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3508, 3486, 3768, 3771, 2641} \[ \frac {2 d^2 \left (7 a^2-2 b^2\right ) \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 d \left (7 a^2-2 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{3/2}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2 - 2*b^2)*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f) + (18*a*b*(d*
Sec[e + f*x])^(5/2))/(35*f) + (2*(7*a^2 - 2*b^2)*d*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(21*f) + (2*b*(d*Sec[e
+ f*x])^(5/2)*(a + b*Tan[e + f*x]))/(7*f)

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 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^2 \, dx &=\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {2}{7} \int (d \sec (e+f x))^{5/2} \left (\frac {7 a^2}{2}-b^2+\frac {9}{2} a b \tan (e+f x)\right ) \, dx\\ &=\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{7} \left (7 a^2-2 b^2\right ) \int (d \sec (e+f x))^{5/2} \, dx\\ &=\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{21} \left (\left (7 a^2-2 b^2\right ) d^2\right ) \int \sqrt {d \sec (e+f x)} \, dx\\ &=\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}+\frac {1}{21} \left (\left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx\\ &=\frac {2 \left (7 a^2-2 b^2\right ) d^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f}+\frac {18 a b (d \sec (e+f x))^{5/2}}{35 f}+\frac {2 \left (7 a^2-2 b^2\right ) d (d \sec (e+f x))^{3/2} \sin (e+f x)}{21 f}+\frac {2 b (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))}{7 f}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 127, normalized size = 0.89 \[ \frac {2 d^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2 \left (\frac {5}{2} \left (7 a^2-2 b^2\right ) \sin (2 (e+f x))+5 \left (7 a^2-2 b^2\right ) \cos ^{\frac {5}{2}}(e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )+3 b (14 a+5 b \tan (e+f x))\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2*(5*(7*a^2 - 2*b^2)*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2
, 2] + (5*(7*a^2 - 2*b^2)*Sin[2*(e + f*x)])/2 + 3*b*(14*a + 5*b*Tan[e + f*x])))/(105*f*(a*Cos[e + f*x] + b*Sin
[e + f*x])^2)

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{2} + 2 \, a b d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right ) + a^{2} d^{2} \sec \left (f x + e\right )^{2}\right )} \sqrt {d \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*d^2*sec(f*x + e)^2*tan(f*x + e)^2 + 2*a*b*d^2*sec(f*x + e)^2*tan(f*x + e) + a^2*d^2*sec(f*x + e)
^2)*sqrt(d*sec(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.95, size = 382, normalized size = 2.67 \[ \frac {2 \left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (35 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}-10 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}+35 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{2}-10 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) b^{2}+35 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{2}-10 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b^{2}+42 \cos \left (f x +e \right ) a b +15 \sin \left (f x +e \right ) b^{2}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{105 f \sin \left (f x +e \right )^{4} \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(35*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*co
s(f*x+e)^4*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^2-10*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*cos(f*x+e)^4*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*b^2+35*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e
)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^3*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^2-10*I*(1/(1+cos(f*x+e)))^(1/
2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^3*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*b^2+35*cos(f*x+e)^
2*sin(f*x+e)*a^2-10*cos(f*x+e)^2*sin(f*x+e)*b^2+42*cos(f*x+e)*a*b+15*sin(f*x+e)*b^2)*(d/cos(f*x+e))^(5/2)/sin(
f*x+e)^4/cos(f*x+e)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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