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

Optimal. Leaf size=113 \[ \frac {3 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac {2 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}}-\frac {2 a^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 124, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3957, 3956, 3795, 203} \[ \frac {3 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac {3 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3957

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1))/(b*f*(2*m +
1)), x] - Dist[(d*(2*n - 1))/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x]
)^(n - 1), 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] && L
tQ[m, -2^(-1)] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx &=-\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac {(3 a) \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx}{2 c}\\ &=-\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac {3 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}}-\frac {\left (3 a^2\right ) \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{c}\\ &=-\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac {3 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}}+\frac {\left (6 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 )}{c f}\\ &=\frac {3 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{c^{3/2} f}-\frac {\left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{f (c-c \sec (e+f x))^{3/2}}-\frac {3 a^2 \tan (e+f x)}{c f \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.53, size = 184, normalized size = 1.63 \[ \frac {a^2 e^{-2 i (e+f x)} \sin \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (4 \left (1+e^{3 i (e+f x)}\right )-3 \sqrt {2} \left (-1+e^{i (e+f x)}\right )^2 \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 c f (\sec (e+f x)-1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 372, normalized size = 3.29 \[ \left [\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \sqrt {-\frac {1}{c}} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} + {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, -\frac {\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right )}{\sqrt {c}} - 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

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))^(3/2),x, algorithm="fricas")

[Out]

[1/2*(3*sqrt(2)*(a^2*c*cos(f*x + e) - a^2*c)*sqrt(-1/c)*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) + 4*(2*a^2*cos(f*x + e)^2 + a^2*cos(f*x + e) - a^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))
)/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e)), -(3*sqrt(2)*(a^2*c*cos(f*x + e) - a^2*c)*arctan(sqrt(2)*sqrt((c
*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e)/sqrt(c) - 2*(2*a^2*cos(f*x
+ e)^2 + a^2*cos(f*x + e) - a^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^2*f*cos(f*x + e) - c^2*f)*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))^(3/2),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)-2/f*2*a^2*(1/2*(-3*(c*tan((f*x+exp(1))/2)^2-c)-2*c)/sqrt(2)/c/(sqrt(c*tan((f*x+exp(
1))/2)^2-c)*(c*tan((f*x+exp(1))/2)^2-c)+c*sqrt(c*tan((f*x+exp(1))/2)^2-c))/sign(tan((f*x+exp(1))/2))/sign(tan(
(f*x+exp(1))/2)^2-1)-3/2*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(c))/sqrt(2)/sqrt(c)/c/sign(tan((f*x+exp(1))
/2))/sign(tan((f*x+exp(1))/2)^2-1))

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maple [A]  time = 1.70, size = 144, normalized size = 1.27 \[ \frac {a^{2} \left (3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )-3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-4 \cos \left (f x +e \right )+2\right ) \sin \left (f x +e \right )}{f \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

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))^(3/2),x)

[Out]

a^2/f*(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)))^(1/2)*cos(f*x+e)-3*arct
an(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-4*cos(f*x+e)+2)*sin(f*x+e)/cos
(f*x+e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)

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

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))^(3/2),x, algorithm="maxima")

[Out]

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

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

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))^(3/2)),x)

[Out]

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

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

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))**(3/2),x)

[Out]

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

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