3.451 \(\int \frac {\sec ^{\frac {3}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx\)

Optimal. Leaf size=240 \[ -\frac {2 \sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^2 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}-\frac {2 \sec ^{\frac {3}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \]

[Out]

-2*sec(e*x+d)^(3/2)*(c*cos(e*x+d)-a*sin(e*x+d))*(b+a*cos(e*x+d)+c*sin(e*x+d))/(a^2-b^2+c^2)/e/(a+b*sec(e*x+d)+
c*tan(e*x+d))^(3/2)-2*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))*Elliptic
E(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*sec(e*x+d)^(3/2)*(b+
a*cos(e*x+d)+c*sin(e*x+d))^2/(a^2-b^2+c^2)/e/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)/(a+b*se
c(e*x+d)+c*tan(e*x+d))^(3/2)

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Rubi [A]  time = 0.22, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3167, 3128, 3119, 2653} \[ -\frac {2 \sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^2 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}-\frac {2 \sec ^{\frac {3}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[d + e*x]^(3/2)*(c*Cos[d + e*x] - a*Sin[d + e*x])*(b + a*Cos[d + e*x] + c*Sin[d + e*x]))/((a^2 - b^2 +
c^2)*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (2*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c
^2])/(b + Sqrt[a^2 + c^2])]*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2)/((a^2 - b^2 + c^2)*e*S
qrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])]*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))

Rule 2653

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

Rule 3119

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

Rule 3128

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

Rule 3167

Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_)
, x_Symbol] :> Dist[(Sec[d + e*x]^n*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n)/(a + b*Sec[d + e*x] + c*Tan[d + e
*x])^n, Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + n, 0] &&
!IntegerQ[n]

Rubi steps

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

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Mathematica [C]  time = 6.40, size = 1732, normalized size = 7.22 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2*((-2*(a^2 + c^2))/(a*c*(a^2 - b^2 + c^2)) + (2*(b*
c + a^2*Sin[d + e*x] + c^2*Sin[d + e*x]))/(a*(a^2 - b^2 + c^2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x]))))/(e*(a
+ b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (2*b*AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin[d
+ e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d
+ e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x]^(3/2)*Sec[d + e*x +
ArcTan[a/c]]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^
2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + A
rcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[
(a^2 + c^2)/c^2])])/(Sqrt[1 + a^2/c^2]*c*(a^2 - b^2 + c^2)*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (a
^2*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b +
a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b +
a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d +
e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x
- ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c^2)/a^2])]*Sqrt[b + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]]]*Sq
rt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2
])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/a]
])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]]))/(c*(a^2 - b^2 + c^2)*e*(a
+ b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)) - (c*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)
*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/
a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a
^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2
)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c^2)/a^2])]*Sqrt[b + a*Sqrt[(a
^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x
- ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]
))/(a^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*
x - ArcTan[c/a]]]))/((a^2 - b^2 + c^2)*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))

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fricas [F]  time = 1.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sec \left (e x + d\right )^{\frac {3}{2}}}{b^{2} \sec \left (e x + d\right )^{2} + c^{2} \tan \left (e x + d\right )^{2} + 2 \, a b \sec \left (e x + d\right ) + a^{2} + 2 \, {\left (b c \sec \left (e x + d\right ) + a c\right )} \tan \left (e x + d\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(e*x+d)^(3/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)*sec(e*x + d)^(3/2)/(b^2*sec(e*x + d)^2 + c^2*tan(e*x + d)^2
+ 2*a*b*sec(e*x + d) + a^2 + 2*(b*c*sec(e*x + d) + a*c)*tan(e*x + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.22, size = 12574, normalized size = 52.39 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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