3.3.1 \(\int \frac {(a+b \sec (e+f x))^{3/2}}{c+d \sec (e+f x)} \, dx\) [201]

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

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 326, normalized size of antiderivative = 1.00, 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 = {4013, 4006, 3869, 3917, 4058} \begin {gather*} -\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};\text {ArcSin}\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};\text {ArcSin}\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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d f} \end {gather*}

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 3869

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

Rule 3917

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

Rule 4006

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 4013

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 4058

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]/(f*(c + d)*Sqrt[a + b*Csc[e + f*x]]*Sqrt[-Cot[e + f*x]^2]))*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))], 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*}

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Mathematica [A]
time = 8.81, size = 230, normalized size = 0.71 \begin {gather*} -\frac {4 \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {\frac {b+a \cos (e+f x)}{(a+b) (1+\cos (e+f x))}} \left ((a-b)^2 c (c+d) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )-2 \left (a^2 \left (c^2-d^2\right ) \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )+(b c-a d)^2 \Pi \left (\frac {c-d}{c+d};\text {ArcSin}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )\right )\right ) \sqrt {a+b \sec (e+f x)}}{c (c-d) (c+d) f (b+a \cos (e+f x))} \end {gather*}

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)*Elliptic
Pi[-1, ArcSin[Tan[(e + f*x)/2]], (a - b)/(a + b)] + (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 = 2.87, size = 581, normalized size = 1.78

method	result	size
default	\(-\frac {2 \sqrt {\frac {a \cos \left (f x +e \right )+b}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {a \cos \left (f x +e \right )+b}{\left (\cos \left (f x +e \right )+1\right ) \left (a +b \right )}}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c^{2}+\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c d -2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a b \,c^{2}-2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a b c d +\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c^{2}+\EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c d -2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} c^{2}+2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} d^{2}-2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} d^{2}+4 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) a b c d -2 \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} c^{2}\right ) \left (-1+\cos \left (f x +e \right )\right )}{f \left (a \cos \left (f x +e \right )+b \right ) \sin \left (f x +e \right )^{2} c \left (c +d \right ) \left (c -d \right )}\)	\(581\)

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,method=_RETURNVERBOSE)

[Out]

-2/f*((a*cos(f*x+e)+b)/cos(f*x+e))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a
+b))^(1/2)*(cos(f*x+e)+1)^2*(EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*c^2+EllipticF((-1+c
os(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+cos(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*EllipticP
i((-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))/sin(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*Elli
pticPi((-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/c/(c+d)/(c-d)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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