3.75 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{\sqrt {c-c \sec (e+f x)}} \, dx\)

Optimal. Leaf size=117 \[ -\frac {4 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 c f}+\frac {16 a^2 \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}} \]

[Out]

-4*a^2*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))*2^(1/2)/f/c^(1/2)+16/3*a^2*tan(f*x+e)/f/(
c-c*sec(f*x+e))^(1/2)-2/3*a^2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/c/f

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Rubi [A]  time = 0.21, antiderivative size = 123, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3956, 3795, 203} \[ -\frac {4 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {4 a^2 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2]*a^2*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(Sqrt[c]*f) + (4*a^2*Tan[e
+ f*x])/(f*Sqrt[c - c*Sec[e + f*x]]) + (2*(a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt[c - c*Sec[e + f*x]]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

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

Rule 3956

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{\sqrt {c-c \sec (e+f x)}} \, dx &=\frac {2 \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}+(2 a) \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx\\ &=\frac {4 a^2 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}+\left (4 a^2\right ) \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx\\ &=\frac {4 a^2 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}-\frac {\left (8 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 c+x^2} \, dx,x,\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}\right )}{f}\\ &=-\frac {4 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {4 a^2 \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}+\frac {2 \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{3 f \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.93, size = 173, normalized size = 1.48 \[ \frac {4 a^2 e^{-\frac {1}{2} i (e+f x)} \sin \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right ) (\sec (e+f x)+7)-3 \sqrt {2} e^{-\frac {1}{2} i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )\right )}{3 f \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*Sec[e + f*x]*((-3*Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 +
E^((2*I)*(e + f*x))])])/E^((I/2)*(e + f*x)) + Cos[(e + f*x)/2]*(7 + Sec[e + f*x]))*(Cos[(e + f*x)/2] + I*Sin[
(e + f*x)/2])*Sin[(e + f*x)/2])/(3*E^((I/2)*(e + f*x))*f*Sqrt[c - c*Sec[e + f*x]])

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fricas [A]  time = 0.53, size = 343, normalized size = 2.93 \[ \left [\frac {2 \, {\left (3 \, \sqrt {2} a^{2} c \sqrt {-\frac {1}{c}} \cos \left (f x + e\right ) \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} - {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - {\left (7 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{3 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\right )}, \frac {2 \, {\left (6 \, \sqrt {2} a^{2} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (7 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{3 \, c f \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/3*(3*sqrt(2)*a^2*c*sqrt(-1/c)*cos(f*x + e)*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt((c*cos(f*x
+ e) - c)/cos(f*x + e))*sqrt(-1/c) - (3*cos(f*x + e) + 1)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin
(f*x + e) - (7*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + a^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*co
s(f*x + e)*sin(f*x + e)), 2/3*(6*sqrt(2)*a^2*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*co
s(f*x + e)/(sqrt(c)*sin(f*x + e)))*cos(f*x + e)*sin(f*x + e) - (7*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + a^
2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*cos(f*x + e)*sin(f*x + e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*((12*i*a^2*atan(-i)-16*a^2)/3/sqrt(-
2*c)*sign(tan((f*x+exp(1))/2))-4*a^2*(1/2*sqrt(2)*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(c))/sqrt(c)/sign(t
an((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)+1/3*(3*(c*tan((f*x+exp(1))/2)^2-c)-c)/sqrt(c*tan((f*x+exp(1)
)/2)^2-c)/(c*tan((f*x+exp(1))/2)^2-c)/sqrt(2)/sign(tan((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)))

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maple [A]  time = 1.75, size = 145, normalized size = 1.24 \[ \frac {2 a^{2} \left (3 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+7 \cos \left (f x +e \right )+1\right ) \sin \left (f x +e \right )}{3 f \cos \left (f x +e \right )^{2} \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a^2/f*(3*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+3*
arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+7*cos(f*x+e)+1)*sin(f*x+e)
/cos(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**(1/2),x)

[Out]

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

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