3.280 \(\int \frac {(g \sec (e+f x))^{3/2} \sqrt {a+b \sec (e+f x)}}{c+d \sec (e+f x)} \, dx\)

Optimal. Leaf size=170 \[ \frac {2 b g \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{d f \sqrt {a+b \sec (e+f x)}}-\frac {2 g (b c-a d) \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (\frac {2 c}{c+d};\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{d f (c+d) \sqrt {a+b \sec (e+f x)}} \]

[Out]

2*b*g*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticPi(sin(1/2*e+1/2*f*x),2,2^(1/2)*(a/(a+b))^(1/2))
*((b+a*cos(f*x+e))/(a+b))^(1/2)*(g*sec(f*x+e))^(1/2)/d/f/(a+b*sec(f*x+e))^(1/2)-2*(-a*d+b*c)*g*(cos(1/2*e+1/2*
f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticPi(sin(1/2*e+1/2*f*x),2*c/(c+d),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(f*
x+e))/(a+b))^(1/2)*(g*sec(f*x+e))^(1/2)/d/(c+d)/f/(a+b*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.84, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {3971, 3859, 2807, 2805, 3975} \[ \frac {2 b g \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{d f \sqrt {a+b \sec (e+f x)}}-\frac {2 g (b c-a d) \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (\frac {2 c}{c+d};\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{d f (c+d) \sqrt {a+b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[((g*Sec[e + f*x])^(3/2)*Sqrt[a + b*Sec[e + f*x]])/(c + d*Sec[e + f*x]),x]

[Out]

(2*b*g*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[2, (e + f*x)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*f
*Sqrt[a + b*Sec[e + f*x]]) - (2*(b*c - a*d)*g*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[(2*c)/(c + d), (e
+ f*x)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*(c + d)*f*Sqrt[a + b*Sec[e + f*x]])

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3971

Int[((csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_)), x_Symbol] :> Dist[b/d, Int[(g*Csc[e + f*x])^(3/2)/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*
c - a*d)/d, Int[(g*Csc[e + f*x])^(3/2)/(Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])), x], x] /; FreeQ[{a, b,
c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3975

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> Dist[(g*Sqrt[g*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]],
Int[1/(Sqrt[b + a*Sin[e + f*x]]*(d + c*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(g \sec (e+f x))^{3/2} \sqrt {a+b \sec (e+f x)}}{c+d \sec (e+f x)} \, dx &=\frac {b \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)}} \, dx}{d}-\frac {(b c-a d) \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx}{d}\\ &=\frac {\left (b g \sqrt {b+a \cos (e+f x)} \sqrt {g \sec (e+f x)}\right ) \int \frac {\sec (e+f x)}{\sqrt {b+a \cos (e+f x)}} \, dx}{d \sqrt {a+b \sec (e+f x)}}-\frac {\left ((b c-a d) g \sqrt {b+a \cos (e+f x)} \sqrt {g \sec (e+f x)}\right ) \int \frac {1}{\sqrt {b+a \cos (e+f x)} (d+c \cos (e+f x))} \, dx}{d \sqrt {a+b \sec (e+f x)}}\\ &=\frac {\left (b g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \sqrt {g \sec (e+f x)}\right ) \int \frac {\sec (e+f x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (e+f x)}{a+b}}} \, dx}{d \sqrt {a+b \sec (e+f x)}}-\frac {\left ((b c-a d) g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \sqrt {g \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (e+f x)}{a+b}} (d+c \cos (e+f x))} \, dx}{d \sqrt {a+b \sec (e+f x)}}\\ &=\frac {2 b g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \Pi \left (2;\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{d f \sqrt {a+b \sec (e+f x)}}-\frac {2 (b c-a d) g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \Pi \left (\frac {2 c}{c+d};\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{d (c+d) f \sqrt {a+b \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 3.98, size = 223, normalized size = 1.31 \[ -\frac {2 i g \cot (e+f x) \sqrt {g \sec (e+f x)} \sqrt {-\frac {a (\cos (e+f x)-1)}{a+b}} \sqrt {\frac {a (\cos (e+f x)+1)}{a-b}} \sqrt {a+b \sec (e+f x)} \left (\Pi \left (1-\frac {a}{b};i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (e+f x)}\right )|\frac {b-a}{a+b}\right )-\Pi \left (\frac {(a-b) c}{a d-b c};i \sinh ^{-1}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (e+f x)}\right )|\frac {b-a}{a+b}\right )\right )}{d f \sqrt {\frac {1}{a-b}} \sqrt {a \cos (e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g*Sec[e + f*x])^(3/2)*Sqrt[a + b*Sec[e + f*x]])/(c + d*Sec[e + f*x]),x]

[Out]

((-2*I)*g*Sqrt[-((a*(-1 + Cos[e + f*x]))/(a + b))]*Sqrt[(a*(1 + Cos[e + f*x]))/(a - b)]*Cot[e + f*x]*(Elliptic
Pi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[e + f*x]]], (-a + b)/(a + b)] - EllipticPi[((a - b)*c)
/(-(b*c) + a*d), I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[e + f*x]]], (-a + b)/(a + b)])*Sqrt[g*Sec[e + f*x
]]*Sqrt[a + b*Sec[e + f*x]])/(Sqrt[(a - b)^(-1)]*d*f*Sqrt[b + a*Cos[e + f*x]])

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fricas [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((g*sec(f*x+e))^(3/2)*(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (f x + e\right ) + a} \left (g \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(3/2)*(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e) + a)*(g*sec(f*x + e))^(3/2)/(d*sec(f*x + e) + c), x)

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maple [C]  time = 2.03, size = 465, normalized size = 2.74 \[ -\frac {2 i \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \left (\EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) a c d +\EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) a \,d^{2}-\EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) b c d -\EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right ) b \,d^{2}-2 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -\frac {c -d}{c +d}, i \sqrt {\frac {a -b}{a +b}}\right ) a c d +2 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -\frac {c -d}{c +d}, i \sqrt {\frac {a -b}{a +b}}\right ) b \,c^{2}-2 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -1, i \sqrt {\frac {a -b}{a +b}}\right ) b \,c^{2}+2 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -1, i \sqrt {\frac {a -b}{a +b}}\right ) b \,d^{2}\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right )}{f \left (b +a \cos \left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, d \left (c +d \right ) \left (c -d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(3/2)*(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x)

[Out]

-2*I/f*((b+a*cos(f*x+e))/(1+cos(f*x+e))/(a+b))^(1/2)*(EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1
/2))*a*c*d+EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))*a*d^2-EllipticF(I*(-1+cos(f*x+e))/sin(
f*x+e),(-(a-b)/(a+b))^(1/2))*b*c*d-EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))*b*d^2-2*Ellipt
icPi(I*(-1+cos(f*x+e))/sin(f*x+e),-(c-d)/(c+d),I*((a-b)/(a+b))^(1/2))*a*c*d+2*EllipticPi(I*(-1+cos(f*x+e))/sin
(f*x+e),-(c-d)/(c+d),I*((a-b)/(a+b))^(1/2))*b*c^2-2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))
^(1/2))*b*c^2+2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))^(1/2))*b*d^2)*((b+a*cos(f*x+e))/cos
(f*x+e))^(1/2)*(g/cos(f*x+e))^(3/2)*cos(f*x+e)^2/(b+a*cos(f*x+e))/(1/(1+cos(f*x+e)))^(1/2)/d/(c+d)/(c-d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sec \left (f x + e\right ) + a} \left (g \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{d \sec \left (f x + e\right ) + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(3/2)*(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e) + a)*(g*sec(f*x + e))^(3/2)/(d*sec(f*x + e) + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/cos(e + f*x))^(1/2)*(g/cos(e + f*x))^(3/2))/(c + d/cos(e + f*x)),x)

[Out]

int(((a + b/cos(e + f*x))^(1/2)*(g/cos(e + f*x))^(3/2))/(c + d/cos(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \sec {\left (e + f x \right )}\right )^{\frac {3}{2}} \sqrt {a + b \sec {\left (e + f x \right )}}}{c + d \sec {\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(3/2)*(a+b*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e)),x)

[Out]

Integral((g*sec(e + f*x))**(3/2)*sqrt(a + b*sec(e + f*x))/(c + d*sec(e + f*x)), x)

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