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

Optimal. Leaf size=149 \[ \frac {2 \sqrt {a} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{d f}-\frac {2 \sqrt {a} \sqrt {c} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \tan (e+f x)}{\sqrt {c+d} \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{d f \sqrt {c+d}} \]

[Out]

2*g^(3/2)*arctanh(a^(1/2)*g^(1/2)*tan(f*x+e)/(g*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)/d/f-2*g^(3/2
)*arctanh(a^(1/2)*c^(1/2)*g^(1/2)*tan(f*x+e)/(c+d)^(1/2)/(g*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)*
c^(1/2)/d/f/(c+d)^(1/2)

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Rubi [A]  time = 0.57, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3970, 3802, 208, 3965} \[ \frac {2 \sqrt {a} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{d f}-\frac {2 \sqrt {a} \sqrt {c} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \tan (e+f x)}{\sqrt {c+d} \sqrt {a \sec (e+f x)+a} \sqrt {g \sec (e+f x)}}\right )}{d f \sqrt {c+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*g^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[g]*Tan[e + f*x])/(Sqrt[g*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])])/(d
*f) - (2*Sqrt[a]*Sqrt[c]*g^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[c]*Sqrt[g]*Tan[e + f*x])/(Sqrt[c + d]*Sqrt[g*Sec[e + f*
x]]*Sqrt[a + a*Sec[e + f*x]])])/(d*Sqrt[c + d]*f)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3802

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/f, Subst[Int[1/(b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x] /; F
reeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] &&  !GtQ[(a*d)/b, 0]

Rule 3965

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

Rule 3970

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

Rubi steps

\begin {align*} \int \frac {(g \sec (e+f x))^{3/2} \sqrt {a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx &=\frac {g \int \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)} \, dx}{d}-\frac {(c g) \int \frac {\sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{d}\\ &=-\frac {\left (2 a g^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-g x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{d f}+\frac {\left (2 a c g^2\right ) \operatorname {Subst}\left (\int \frac {1}{a c+a d-c g x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{d f}\\ &=\frac {2 \sqrt {a} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{d f}-\frac {2 \sqrt {a} \sqrt {c} g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \tan (e+f x)}{\sqrt {c+d} \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{d \sqrt {c+d} f}\\ \end {align*}

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Mathematica [C]  time = 1.37, size = 427, normalized size = 2.87 \[ \frac {\left (\sqrt {2}-2 i\right ) g^2 \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (i \left (2 \sqrt {c+d} \log \left (2 \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {2}\right )+2 \sqrt {c} \log \left (\sqrt {2} \sqrt {c+d}-2 \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )\right )-2 \sqrt {c} \log \left (\sqrt {2} \sqrt {c+d}+2 \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )\right )-\sqrt {c+d} \log \left (-\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )-\sqrt {2} \cos \left (\frac {1}{2} (e+f x)\right )+2\right )-\sqrt {c+d} \log \left (-\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {2} \cos \left (\frac {1}{2} (e+f x)\right )+2\right )\right )+2 \sqrt {c+d} \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (e+f x)\right )-\left (\sqrt {2}-1\right ) \sin \left (\frac {1}{4} (e+f x)\right )}{\left (1+\sqrt {2}\right ) \cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )}\right )+2 \sqrt {c+d} \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (e+f x)\right )-\left (1+\sqrt {2}\right ) \sin \left (\frac {1}{4} (e+f x)\right )}{\left (\sqrt {2}-1\right ) \cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )}\right )\right )}{4 \left (\sqrt {2}+i\right ) d f \sqrt {c+d} \sqrt {g \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I + Sqrt[2])*g^2*(2*Sqrt[c + d]*ArcTan[(Cos[(e + f*x)/4] - (-1 + Sqrt[2])*Sin[(e + f*x)/4])/((1 + Sqrt[2]
)*Cos[(e + f*x)/4] - Sin[(e + f*x)/4])] + 2*Sqrt[c + d]*ArcTan[(Cos[(e + f*x)/4] - (1 + Sqrt[2])*Sin[(e + f*x)
/4])/((-1 + Sqrt[2])*Cos[(e + f*x)/4] - Sin[(e + f*x)/4])] + I*(2*Sqrt[c + d]*Log[Sqrt[2] + 2*Sin[(e + f*x)/2]
] - Sqrt[c + d]*Log[2 - Sqrt[2]*Cos[(e + f*x)/2] - Sqrt[2]*Sin[(e + f*x)/2]] - Sqrt[c + d]*Log[2 + Sqrt[2]*Cos
[(e + f*x)/2] - Sqrt[2]*Sin[(e + f*x)/2]] + 2*Sqrt[c]*Log[Sqrt[2]*Sqrt[c + d] - 2*Sqrt[c]*Sin[(e + f*x)/2]] -
2*Sqrt[c]*Log[Sqrt[2]*Sqrt[c + d] + 2*Sqrt[c]*Sin[(e + f*x)/2]]))*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])
/(4*(I + Sqrt[2])*d*Sqrt[c + d]*f*Sqrt[g*Sec[e + f*x]])

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fricas [A]  time = 7.93, size = 1126, normalized size = 7.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a*c*g/(c + d))*g*log((a*c^2*g*cos(f*x + e)^3 - (7*a*c^2 + 6*a*c*d)*g*cos(f*x + e)^2 + 4*((c^2 + c*d
)*cos(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(a*c*g/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x +
e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*a*c*d + a*d^2)*g*cos(f*x + e) + (8*a*c^2 + 8*a*c*d + a*d^2)*g)/(c^
2*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))) + sqrt(a*g)*g*log((a*g*co
s(f*x + e)^3 - 7*a*g*cos(f*x + e)^2 - 4*sqrt(a*g)*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/
cos(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + 8*a*g)/(cos(f*x + e)^3 + cos(f*x + e)^2)))/(d*f), -1/2*(2*sq
rt(-a*c*g/(c + d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-a*c*g/(c + d))*sqrt((a*cos(f
*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(a*c*g*sin(f*x + e))) - sqrt(a*g)*g*log((a*g*cos(f*x + e)^3 -
7*a*g*cos(f*x + e)^2 - 4*sqrt(a*g)*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*s
qrt(g/cos(f*x + e))*sin(f*x + e) + 8*a*g)/(cos(f*x + e)^3 + cos(f*x + e)^2)))/(d*f), 1/2*(2*sqrt(-a*g)*g*arcta
n(2*sqrt(-a*g)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(a*g*cos
(f*x + e)^2 - a*g*cos(f*x + e) - 2*a*g)) + sqrt(a*c*g/(c + d))*g*log((a*c^2*g*cos(f*x + e)^3 - (7*a*c^2 + 6*a*
c*d)*g*cos(f*x + e)^2 + 4*((c^2 + c*d)*cos(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(a*c*g/(c + d)
)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*a*c*d + a*d^2)*g*cos(f*x + e)
+ (8*a*c^2 + 8*a*c*d + a*d^2)*g)/(c^2*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos
(f*x + e))))/(d*f), (sqrt(-a*g)*g*arctan(2*sqrt(-a*g)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x +
e))*cos(f*x + e)*sin(f*x + e)/(a*g*cos(f*x + e)^2 - a*g*cos(f*x + e) - 2*a*g)) - sqrt(-a*c*g/(c + d))*g*arcta
n(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-a*c*g/(c + d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))
*sqrt(g/cos(f*x + e))/(a*c*g*sin(f*x + e))))/(d*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.17, size = 566, normalized size = 3.80 \[ \frac {2 \left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) \left (-1+\cos \left (f x +e \right )\right )^{2} \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (\sqrt {\frac {c}{c -d}}\, \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right )}{2}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}+\sqrt {\frac {c}{c -d}}\, \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (-\cos \left (f x +e \right )-1+\sin \left (f x +e \right )\right )}{2}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}+c \ln \left (\frac {4 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, c \sin \left (f x +e \right )-4 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, d \sin \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-2 \sqrt {\left (c +d \right ) \left (c -d \right )}}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d}\right )-c \ln \left (-\frac {2 \left (2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, c \sin \left (f x +e \right )-2 \sqrt {\frac {c}{c -d}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, d \sin \left (f x +e \right )+c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )+c \cos \left (f x +e \right )-d \cos \left (f x +e \right )-c +d}\right )\right ) \left (c -d \right )}{f \sin \left (f x +e \right )^{4} \left (\frac {1}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) \left (c -d \right )}\, \left (c -d +\sqrt {\left (c +d \right ) \left (c -d \right )}\right ) \left (-c +d +\sqrt {\left (c +d \right ) \left (c -d \right )}\right ) \sqrt {\frac {c}{c -d}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x)

[Out]

2/f*(g/cos(f*x+e))^(3/2)*cos(f*x+e)^2*(-1+cos(f*x+e))^2*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*((c/(c-d))^(1/2)*a
rctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))*((c+d)*(c-d))^(1/2)+(c/(c-d))^(1/2)*arctanh(1/2
*(1/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)-1+sin(f*x+e)))*((c+d)*(c-d))^(1/2)+c*ln(2*(2*(c/(c-d))^(1/2)*(1/(1+cos(
f*x+e)))^(1/2)*c*sin(f*x+e)-2*(c/(c-d))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*d*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+
((c+d)*(c-d))^(1/2)*cos(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+
c-d))-c*ln(-2*(2*(c/(c-d))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*c*sin(f*x+e)-2*(c/(c-d))^(1/2)*(1/(1+cos(f*x+e)))^(1
/2)*d*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2)*cos(f*x+e)+((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^
(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)))/sin(f*x+e)^4/(1/(1+cos(f*x+e)))^(3/2)*(c-d)/((c+d)*(c-d))^(1
/2)/(c-d+((c+d)*(c-d))^(1/2))/(-c+d+((c+d)*(c-d))^(1/2))/(c/(c-d))^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (g \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{c + d \sec {\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(3/2)*(a+a*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e)),x)

[Out]

Integral(sqrt(a*(sec(e + f*x) + 1))*(g*sec(e + f*x))**(3/2)/(c + d*sec(e + f*x)), x)

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