3.6.88 \(\int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2 \, dx\) [588]

Optimal. Leaf size=103 \[ \frac {10 a b \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}{3 f} \]

[Out]

10/3*a*b*(d*sec(f*x+e))^(1/2)/f+2/3*(3*a^2-2*b^2)*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(si
n(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(d*sec(f*x+e))^(1/2)/f+2/3*b*(d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))/
f

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Rubi [A]
time = 0.09, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3589, 3567, 3856, 2720} \begin {gather*} \frac {2 \left (3 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)}}{3 f}+\frac {10 a b \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*a*b*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*(3*a^2 - 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*S
ec[e + f*x]])/(3*f) + (2*b*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x]))/(3*f)

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

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Mathematica [A]
time = 0.70, size = 87, normalized size = 0.84 \begin {gather*} \frac {2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (\left (3 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 )+b \cos (e+f x) (6 a \cos (e+f x)+b \sin (e+f x))\right )}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.75, size = 339, normalized size = 3.29

method	result	size
default	\(\frac {2 \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )-1\right )^{2} \left (3 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \left (\cos ^{2}\left (f x +e \right )\right ) a^{2}-2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \left (\cos ^{2}\left (f x +e \right )\right ) b^{2}+3 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a^{2}-2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) b^{2}+6 \cos \left (f x +e \right ) a b +\sin \left (f x +e \right ) b^{2}\right ) \left (\cos \left (f x +e \right )+1\right )^{2}}{3 f \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4}}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/3/f*(d/cos(f*x+e))^(1/2)*(cos(f*x+e)-1)^2*(3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*El
lipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*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)*cos(f*x+e)^2*b^2+3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*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)*cos(f*x+e)*b^2+6*cos(f*x+e)*a*b+sin(f*x+
e)*b^2)*(cos(f*x+e)+1)^2/cos(f*x+e)/sin(f*x+e)^4

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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((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 143, normalized size = 1.39 \begin {gather*} \frac {\sqrt {2} {\left (-3 i \, a^{2} + 2 i \, b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2} {\left (3 i \, a^{2} - 2 i \, b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\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 ) + b^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/2)*(a+b*tan(f*x+e))**2,x)

[Out]

Integral(sqrt(d*sec(e + f*x))*(a + b*tan(e + f*x))**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((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))^2,x)

[Out]

int((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))^2, x)

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