3.6.79 \(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx\) [579]

Optimal. Leaf size=92 \[ \frac {2 a d^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \]

[Out]

2/5*b*(d*sec(f*x+e))^(5/2)/f+2/3*a*d*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/3*a*d^2*(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)/f

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Rubi [A]
time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3567, 3853, 3856, 2720} \begin {gather*} \frac {2 a 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)}}{3 f}+\frac {2 a d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*d^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*b*(d*Sec[e + f*x])^(5/2
))/(5*f) + (2*a*d*(d*Sec[e + f*x])^(3/2)*Sin[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 3853

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

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Mathematica [A]
time = 0.47, size = 58, normalized size = 0.63 \begin {gather*} \frac {(d \sec (e+f x))^{5/2} \left (6 b+10 a \cos ^{\frac {5}{2}}(e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )+5 a \sin (2 (e+f x))\right )}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sec[e + f*x])^(5/2)*(6*b + 10*a*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x)/2, 2] + 5*a*Sin[2*(e + f*x)]))/(15*
f)

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Maple [C] Result contains complex when optimal does not.
time = 0.47, size = 195, normalized size = 2.12

method	result	size
default	\(\frac {2 \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \left (\cos ^{3}\left (f x +e \right )\right ) a +5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \left (\cos ^{2}\left (f x +e \right )\right ) a +5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +3 b \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{15 f \sin \left (f x +e \right )^{4}}\)	\(195\)

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)),x,method=_RETURNVERBOSE)

[Out]

2/15/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(5*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellip
ticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*a+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)*cos(f*x+e)^2*a+5*cos(f*x+e)*sin(f*x+e)*a+3*b)*(d/cos(f*x+e))^(5/2
)/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))^(5/2)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 133, normalized size = 1.45 \begin {gather*} \frac {-5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, a d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, b d^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \end {gather*}

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)),x, algorithm="fricas")

[Out]

1/15*(-5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 5*I*sq
rt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(5*a*d^2*cos(f*x
+ e)*sin(f*x + e) + 3*b*d^2)*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)

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

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)),x)

[Out]

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

[Out]

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

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

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)),x)

[Out]

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

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