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

Optimal. Leaf size=326 \[ \frac{2 b \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{d f}-\frac{2 (b c-a d)^2 \tan (e+f x) \sqrt{\frac{a+b \sec (e+f x)}{a+b}} \Pi \left (\frac{2 d}{c+d};\sin ^{-1}\left (\frac{\sqrt{1-\sec (e+f x)}}{\sqrt{2}}\right )|\frac{2 b}{a+b}\right )}{c d f (c+d) \sqrt{-\tan ^2(e+f x)} \sqrt{a+b \sec (e+f x)}}-\frac{2 a \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{c f} \]

[Out]

(2*b*Sqrt[a + b]*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b
*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(d*f) - (2*a*Sqrt[a + b]*Cot[e + f*x]*E
llipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x])
)/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(c*f) - (2*(b*c - a*d)^2*EllipticPi[(2*d)/(c + d), ArcSin[
Sqrt[1 - Sec[e + f*x]]/Sqrt[2]], (2*b)/(a + b)]*Sqrt[(a + b*Sec[e + f*x])/(a + b)]*Tan[e + f*x])/(c*d*(c + d)*
f*Sqrt[a + b*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^2])

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Rubi [A]  time = 0.361492, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3928, 3921, 3784, 3832, 3973} \[ -\frac{2 (b c-a d)^2 \tan (e+f x) \sqrt{\frac{a+b \sec (e+f x)}{a+b}} \Pi \left (\frac{2 d}{c+d};\sin ^{-1}\left (\frac{\sqrt{1-\sec (e+f x)}}{\sqrt{2}}\right )|\frac{2 b}{a+b}\right )}{c d f (c+d) \sqrt{-\tan ^2(e+f x)} \sqrt{a+b \sec (e+f x)}}-\frac{2 a \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{c f}+\frac{2 b \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[a + b]*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b
*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(d*f) - (2*a*Sqrt[a + b]*Cot[e + f*x]*E
llipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x])
)/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(c*f) - (2*(b*c - a*d)^2*EllipticPi[(2*d)/(c + d), ArcSin[
Sqrt[1 - Sec[e + f*x]]/Sqrt[2]], (2*b)/(a + b)]*Sqrt[(a + b*Sec[e + f*x])/(a + b)]*Tan[e + f*x])/(c*d*(c + d)*
f*Sqrt[a + b*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^2])

Rule 3928

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[1/(
c*d), Int[(a^2*d + b^2*c*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)^2/(c*d), Int[Csc[e
+ f*x]/(Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])), x], 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]

Rule 3921

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

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3973

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
), x_Symbol] :> Simp[(-2*Cot[e + f*x]*Sqrt[(a + b*Csc[e + f*x])/(a + b)]*EllipticPi[(2*d)/(c + d), ArcSin[Sqrt
[1 - Csc[e + f*x]]/Sqrt[2]], (2*b)/(a + b)])/(f*(c + d)*Sqrt[a + b*Csc[e + f*x]]*Sqrt[-Cot[e + f*x]^2]), 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]

Rubi steps

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

Mathematica [A]  time = 5.62718, size = 233, normalized size = 0.71 \[ -\frac{4 \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sqrt{\frac{\cos (e+f x)}{\cos (e+f x)+1}} \sqrt{\frac{a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} \sqrt{a+b \sec (e+f x)} \left (c (a-b)^2 (c+d) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (e+f x)\right )\right ),\frac{a-b}{a+b}\right )+2 a^2 \left (c^2-d^2\right ) \Pi \left (-1;-\sin ^{-1}\left (\tan \left (\frac{1}{2} (e+f x)\right )\right )|\frac{a-b}{a+b}\right )+2 (b c-a d)^2 \Pi \left (\frac{c-d}{c+d};-\sin ^{-1}\left (\tan \left (\frac{1}{2} (e+f x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{c f (c-d) (c+d) (a \cos (e+f x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]*Sqrt[(b + a*Cos[e + f*x])/((a + b)*(1 + Cos[e + f
*x]))]*((a - b)^2*c*(c + d)*EllipticF[ArcSin[Tan[(e + f*x)/2]], (a - b)/(a + b)] + 2*a^2*(c^2 - d^2)*EllipticP
i[-1, -ArcSin[Tan[(e + f*x)/2]], (a - b)/(a + b)] + 2*(b*c - a*d)^2*EllipticPi[(c - d)/(c + d), -ArcSin[Tan[(e
+ f*x)/2]], (a - b)/(a + b)])*Sqrt[a + b*Sec[e + f*x]])/(c*(c - d)*(c + d)*f*(b + a*Cos[e + f*x]))

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Maple [A]  time = 0.305, size = 581, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f/c/(c+d)/(c-d)*(1/cos(f*x+e)*(a*cos(f*x+e)+b))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x
+e)+b)/(1+cos(f*x+e)))^(1/2)*(1+cos(f*x+e))^2*(EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*c
^2+EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*c*d-2*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((
a-b)/(a+b))^(1/2))*a*b*c^2-2*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b*c*d+EllipticF((-1+c
os(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^2*c^2+EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b
^2*c*d-2*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^2*c^2+2*EllipticPi((-1+cos(f*x+e))/si
n(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^2*d^2-2*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),(c-d)/(c+d),((a-b)/(a+b))^(1/
2))*a^2*d^2+4*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),(c-d)/(c+d),((a-b)/(a+b))^(1/2))*a*b*c*d-2*EllipticPi((-1+
cos(f*x+e))/sin(f*x+e),(c-d)/(c+d),((a-b)/(a+b))^(1/2))*b^2*c^2)*(-1+cos(f*x+e))/(a*cos(f*x+e)+b)/sin(f*x+e)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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