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

Optimal. Leaf size=231 \[ -\frac {2 c \cot (e+f x) \sqrt {-\frac {d (1-\sec (e+f x))}{c+d \sec (e+f x)}} \sqrt {\frac {d (\sec (e+f x)+1)}{c+d \sec (e+f x)}} (c+d \sec (e+f x)) \Pi \left (\frac {c}{c+d};\sin ^{-1}\left (\frac {\sqrt {c+d}}{\sqrt {c+d \sec (e+f x)}}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {c+d}}-\frac {(c-d) \sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {c+d \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {\frac {c+d \sec (e+f x)}{(c+d) (\sec (e+f x)+1)}}} \]

[Out]

-2*c*cot(f*x+e)*EllipticPi((c+d)^(1/2)/(c+d*sec(f*x+e))^(1/2),c/(c+d),((c-d)/(c+d))^(1/2))*(c+d*sec(f*x+e))*(-
d*(1-sec(f*x+e))/(c+d*sec(f*x+e)))^(1/2)*(d*(1+sec(f*x+e))/(c+d*sec(f*x+e)))^(1/2)/a/f/(c+d)^(1/2)-(c-d)*Ellip
ticE(tan(f*x+e)/(1+sec(f*x+e)),((c-d)/(c+d))^(1/2))*(1/(1+sec(f*x+e)))^(1/2)*(c+d*sec(f*x+e))^(1/2)/a/f/((c+d*
sec(f*x+e))/(c+d)/(1+sec(f*x+e)))^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3927, 3780, 3968} \[ -\frac {2 c \cot (e+f x) \sqrt {-\frac {d (1-\sec (e+f x))}{c+d \sec (e+f x)}} \sqrt {\frac {d (\sec (e+f x)+1)}{c+d \sec (e+f x)}} (c+d \sec (e+f x)) \Pi \left (\frac {c}{c+d};\sin ^{-1}\left (\frac {\sqrt {c+d}}{\sqrt {c+d \sec (e+f x)}}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {c+d}}-\frac {(c-d) \sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {c+d \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {\frac {c+d \sec (e+f x)}{(c+d) (\sec (e+f x)+1)}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3780

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

Rule 3927

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

Rule 3968

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

Rubi steps

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

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Mathematica [B]  time = 18.23, size = 810, normalized size = 3.51 \[ \frac {(c+d \sec (e+f x))^{3/2} \left (2 \sec \left (\frac {1}{2} (e+f x)\right ) \left (d \sin \left (\frac {1}{2} (e+f x)\right )-c \sin \left (\frac {1}{2} (e+f x)\right )\right )-2 (d-c) \sin (e+f x)\right ) \cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{f (d+c \cos (e+f x)) (\sec (e+f x) a+a)}+\frac {2 (c+d \sec (e+f x))^{3/2} \left (c^2 \tan ^5\left (\frac {1}{2} (e+f x)\right )+d^2 \tan ^5\left (\frac {1}{2} (e+f x)\right )-2 c d \tan ^5\left (\frac {1}{2} (e+f x)\right )-2 c^2 \tan ^3\left (\frac {1}{2} (e+f x)\right )+2 c d \tan ^3\left (\frac {1}{2} (e+f x)\right )-4 c^2 \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {-c \tan ^2\left (\frac {1}{2} (e+f x)\right )+d \tan ^2\left (\frac {1}{2} (e+f x)\right )+c+d}{c+d}} \tan ^2\left (\frac {1}{2} (e+f x)\right )+c^2 \tan \left (\frac {1}{2} (e+f x)\right )-d^2 \tan \left (\frac {1}{2} (e+f x)\right )+\left (c^2-d^2\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right ) \sqrt {\frac {-c \tan ^2\left (\frac {1}{2} (e+f x)\right )+d \tan ^2\left (\frac {1}{2} (e+f x)\right )+c+d}{c+d}}+2 c (c-d) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right ) \sqrt {\frac {-c \tan ^2\left (\frac {1}{2} (e+f x)\right )+d \tan ^2\left (\frac {1}{2} (e+f x)\right )+c+d}{c+d}}-4 c^2 \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {-c \tan ^2\left (\frac {1}{2} (e+f x)\right )+d \tan ^2\left (\frac {1}{2} (e+f x)\right )+c+d}{c+d}}\right ) \cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right )}{f (d+c \cos (e+f x))^{3/2} \sqrt {\sec (e+f x)} (\sec (e+f x) a+a) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (e+f x)\right )}} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right ) \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right )^{3/2} \sqrt {\frac {-c \tan ^2\left (\frac {1}{2} (e+f x)\right )+d \tan ^2\left (\frac {1}{2} (e+f x)\right )+c+d}{\tan ^2\left (\frac {1}{2} (e+f x)\right )+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e/2 + (f*x)/2]^2*(c + d*Sec[e + f*x])^(3/2)*(2*Sec[(e + f*x)/2]*(-(c*Sin[(e + f*x)/2]) + d*Sin[(e + f*x)/
2]) - 2*(-c + d)*Sin[e + f*x]))/(f*(d + c*Cos[e + f*x])*(a + a*Sec[e + f*x])) + (2*Cos[e/2 + (f*x)/2]^2*(c + d
*Sec[e + f*x])^(3/2)*(c^2*Tan[(e + f*x)/2] - d^2*Tan[(e + f*x)/2] - 2*c^2*Tan[(e + f*x)/2]^3 + 2*c*d*Tan[(e +
f*x)/2]^3 + c^2*Tan[(e + f*x)/2]^5 - 2*c*d*Tan[(e + f*x)/2]^5 + d^2*Tan[(e + f*x)/2]^5 - 4*c^2*EllipticPi[-1,
ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(c + d - c*Tan[(e + f*x)/2]^2 + d
*Tan[(e + f*x)/2]^2)/(c + d)] - 4*c^2*EllipticPi[-1, ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)]*Tan[(e + f*x)/
2]^2*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(c + d - c*Tan[(e + f*x)/2]^2 + d*Tan[(e + f*x)/2]^2)/(c + d)] + (c^2 -
d^2)*EllipticE[ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1 + Tan[(e + f*x)/2]^
2)*Sqrt[(c + d - c*Tan[(e + f*x)/2]^2 + d*Tan[(e + f*x)/2]^2)/(c + d)] + 2*c*(c - d)*EllipticF[ArcSin[Tan[(e +
f*x)/2]], (c - d)/(c + d)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(c + d - c*Tan[(e + f*x
)/2]^2 + d*Tan[(e + f*x)/2]^2)/(c + d)]))/(f*(d + c*Cos[e + f*x])^(3/2)*Sqrt[Sec[e + f*x]]*(a + a*Sec[e + f*x]
)*Sqrt[(1 - Tan[(e + f*x)/2]^2)^(-1)]*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e + f*x)/2]^2)^(3/2)*Sqrt[(c + d - c
*Tan[(e + f*x)/2]^2 + d*Tan[(e + f*x)/2]^2)/(1 + Tan[(e + f*x)/2]^2)])

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{a \sec \left (f x + e\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.92, size = 295, normalized size = 1.28 \[ -\frac {\sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \sqrt {\frac {d +c \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (c +d \right )}}\, \left (-1+\cos \left (f x +e \right )\right ) \left (2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) c^{2}-2 c \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) d +\EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) c^{2}-\EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) d^{2}-4 c^{2} \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {c -d}{c +d}}\right )\right )}{a f \left (d +c \cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c \sqrt {c + d \sec {\left (e + f x \right )}}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d \sqrt {c + d \sec {\left (e + f x \right )}} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c*sqrt(c + d*sec(e + f*x))/(sec(e + f*x) + 1), x) + Integral(d*sqrt(c + d*sec(e + f*x))*sec(e + f*x)
/(sec(e + f*x) + 1), x))/a

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