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

Optimal. Leaf size=122 \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} a c^{3/2} f}-\frac {3 \tan (e+f x)}{4 a f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}} \]

[Out]

-3/8*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a/c^(3/2)/f*2^(1/2)-3/4*tan(f*x+e)/a/f/(c-c
*sec(f*x+e))^(3/2)+tan(f*x+e)/f/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^(3/2)

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Rubi [A]  time = 0.21, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 = {3960, 3796, 3795, 203} \[ -\frac {3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} a c^{3/2} f}-\frac {3 \tan (e+f x)}{4 a f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3796

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

Rule 3960

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

Rubi steps

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

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Mathematica [C]  time = 1.42, size = 183, normalized size = 1.50 \[ \frac {e^{-\frac {1}{2} i (e+f x)} \csc (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-8 (\cos (e+f x)-3)+\frac {6 \sqrt {2} e^{-i (e+f x)} \left (-1+e^{i (e+f x)}\right )^2 \left (1+e^{i (e+f x)}\right ) \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )}{\sqrt {1+e^{2 i (e+f x)}}}\right )}{32 a c f \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((6*Sqrt[2]*(-1 + E^(I*(e + f*x)))^2*(1 + E^(I*(e + f*x)))*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^
((2*I)*(e + f*x))])])/(E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]) - 8*(-3 + Cos[e + f*x]))*Csc[e + f*x]*(C
os[(e + f*x)/2] + I*Sin[(e + f*x)/2]))/(32*a*c*E^((I/2)*(e + f*x))*f*Sqrt[c - c*Sec[e + f*x]])

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fricas [A]  time = 0.51, size = 329, normalized size = 2.70 \[ \left [-\frac {3 \, \sqrt {2} \sqrt {-c} {\left (\cos \left (f x + e\right ) - 1\right )} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (\cos \left (f x + e\right )^{2} - 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{16 \, {\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} f\right )} \sin \left (f x + e\right )}, \frac {3 \, \sqrt {2} \sqrt {c} {\left (\cos \left (f x + e\right ) - 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (\cos \left (f x + e\right )^{2} - 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \, {\left (a c^{2} f \cos \left (f x + e\right ) - a c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*sqrt(2)*sqrt(-c)*(cos(f*x + e) - 1)*log((2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*
cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(
f*x + e) + 4*(cos(f*x + e)^2 - 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a*c^2*f*cos(f*x + e)
- a*c^2*f)*sin(f*x + e)), 1/8*(3*sqrt(2)*sqrt(c)*(cos(f*x + e) - 1)*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) - 2*(cos(f*x + e)^2 - 3*cos(f*x + e))*sqrt((c*
cos(f*x + e) - c)/cos(f*x + e)))/((a*c^2*f*cos(f*x + e) - a*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))/(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)2/f/4*(sqrt(c*tan((f*x+exp(1))/2)^2-c)-3/2*sqrt(c)*atan(sqrt(c
*tan((f*x+exp(1))/2)^2-c)/sqrt(c))+1/2*c*sqrt(c*tan((f*x+exp(1))/2)^2-c)/c/tan((f*x+exp(1))/2)^2)/sqrt(2)/c^2/
a/sign(tan((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)

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maple [B]  time = 2.14, size = 266, normalized size = 2.18 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (\left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )+\left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}}+\cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}-\left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}-3 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-3 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )+3 \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{2 a f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{3} \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/f*(-1+cos(f*x+e))^2*((-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2)*cos(f*x+e)+(-2*cos(f*x+e)/(1+cos(f*x+e)))^(5/2
)+cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)-3*cos(f*x+e)*(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(1/2)-3*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))+3*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(1/2)+3*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x
+e)^3/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(cos(e + f*x)*(a + a/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 \[ \frac {\int \frac {\sec {\left (e + f x \right )}}{- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + c \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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