3.456 \(\int \frac{1}{\cos ^{\frac{3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx\)

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

[Out]

(-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*Cos[d + e*x]
^(3/2)*(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])]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2)/((a^2 - b^2 + c^2)*e*Cos[d + e*x]^(3/2)*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))

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Rubi [A]  time = 0.211226, antiderivative size = 240, normalized size of antiderivative = 1., 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 = {3163, 3128, 3119, 2653} \[ -\frac{2 (a \cos (d+e x)+b+c \sin (d+e x))^2 E\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac{2 \sqrt{a^2+c^2}}{b+\sqrt{a^2+c^2}}\right )}{e \left (a^2-b^2+c^2\right ) \cos ^{\frac{3}{2}}(d+e x) \sqrt{\frac{a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt{a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}-\frac{2 (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{e \left (a^2-b^2+c^2\right ) \cos ^{\frac{3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[d + e*x]^(3/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)),x]

[Out]

(-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*Cos[d + e*x]
^(3/2)*(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])]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2)/((a^2 - b^2 + c^2)*e*Cos[d + e*x]^(3/2)*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 3163

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

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 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 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]

Rubi steps

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

Mathematica [F]  time = 24.5685, size = 0, normalized size = 0. \[ \int \frac{1}{\cos ^{\frac{3}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/(Cos[d + e*x]^(3/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)),x]

[Out]

Integrate[1/(Cos[d + e*x]^(3/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)), x]

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Maple [C]  time = 0.44, size = 12562, normalized size = 52.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac{3}{2}} \cos \left (e x + d\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt{\cos \left (e x + d\right )}}{b^{2} \cos \left (e x + d\right )^{2} \sec \left (e x + d\right )^{2} + c^{2} \cos \left (e x + d\right )^{2} \tan \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right )^{2} \sec \left (e x + d\right ) + a^{2} \cos \left (e x + d\right )^{2} + 2 \,{\left (b c \cos \left (e x + d\right )^{2} \sec \left (e x + d\right ) + a c \cos \left (e x + d\right )^{2}\right )} \tan \left (e x + d\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(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)*sqrt(cos(e*x + d))/(b^2*cos(e*x + d)^2*sec(e*x + d)^2 + c^2
*cos(e*x + d)^2*tan(e*x + d)^2 + 2*a*b*cos(e*x + d)^2*sec(e*x + d) + a^2*cos(e*x + d)^2 + 2*(b*c*cos(e*x + d)^
2*sec(e*x + d) + a*c*cos(e*x + d)^2)*tan(e*x + d)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac{3}{2}} \cos \left (e x + d\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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