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

Optimal. Leaf size=137 \[ \frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^3 f}+\frac {4 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a c^3 f}-\frac {2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^3 f}+\frac {2 a \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^3 f} \]

[Out]

2*a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^3/f-2/3*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^3/
f+4/5*cot(f*x+e)^5*(a+a*sec(f*x+e))^(5/2)/a/c^3/f+2*a*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^3/f

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3904, 3887, 453, 325, 203} \[ \frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^3 f}+\frac {4 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a c^3 f}-\frac {2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^3 f}+\frac {2 a \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^3*f) + (2*a*Cot[e + f*x]*Sqrt[a + a*Sec
[e + f*x]])/(c^3*f) - (2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*c^3*f) + (4*Cot[e + f*x]^5*(a + a*Sec[e
+ f*x])^(5/2))/(5*a*c^3*f)

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps

\begin {align*} \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^3} \, dx &=-\frac {\int \cot ^6(e+f x) (a+a \sec (e+f x))^{9/2} \, dx}{a^3 c^3}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {2+a x^2}{x^6 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a c^3 f}\\ &=\frac {4 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^3 f}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}\\ &=-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^3 f}+\frac {4 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^3 f}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}\\ &=\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^3 f}+\frac {4 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^3 f}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}\\ &=\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^3 f}+\frac {4 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^3 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.80, size = 102, normalized size = 0.74 \[ \frac {2 a \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (5 (\cos (e+f x)-1) \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+6 \cos ^{\frac {5}{2}}(e+f x)\right )}{15 c^3 f \cos ^{\frac {5}{2}}(e+f x) (\sec (e+f x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(6*Cos[e + f*x]^(5/2) + 5*(-1 + Cos[e + f*x])*Hypergeometric2F1[-3/2, -3/2, -1/2, 2*Sin[(e + f*x)/2]^2])*
Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(15*c^3*f*Cos[e + f*x]^(5/2)*(-1 + Sec[e + f*x])^3)

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 417, normalized size = 3.04 \[ \left [\frac {15 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (26 \, a \cos \left (f x + e\right )^{3} - 35 \, a \cos \left (f x + e\right )^{2} + 15 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{30 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (26 \, a \cos \left (f x + e\right )^{3} - 35 \, a \cos \left (f x + e\right )^{2} + 15 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(-a)*log(-(8*a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 -
cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x +
e) + 1))*sin(f*x + e) + 4*(26*a*cos(f*x + e)^3 - 35*a*cos(f*x + e)^2 + 15*a*cos(f*x + e))*sqrt((a*cos(f*x + e
) + a)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)), 1/15*(15*(a*cos(f*
x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)
*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 2*(26*a*cos(f*x + e)^3 - 35*a*cos(f*x
+ e)^2 + 15*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x
+ e) + c^3*f)*sin(f*x + e))]

________________________________________________________________________________________

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*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)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)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)-4*(1/4123168604160*(5222680231936*a^6*s
qrt(-a)*sign(cos(f*x+exp(1)))[phone]*a^2*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*
(f*x+exp(1))))^8*sign(cos(f*x+exp(1)))-20615843020800*a^3*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a
)*tan(1/2*(f*x+exp(1))))^6*sign(cos(f*x+exp(1)))[phone]*a^4*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+
a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^4*sign(cos(f*x+exp(1)))-17867063951360*a^5*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+e
xp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2*sign(cos(f*x+exp(1))))/sqrt(2)/c^3/((sqrt(-a*tan(1/2*(f*x+exp(1
)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-a)^5+1/4*a^2*sqrt(-a)*sign(cos(f*x+exp(1)))*ln(abs(2*(sqrt(-a*tan(1
/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-4*sqrt(2)*abs(a)-6*a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(
1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+4*sqrt(2)*abs(a)-6*a))/c^3/abs(a))/f

________________________________________________________________________________________

maple [B]  time = 1.76, size = 304, normalized size = 2.22 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (-15 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )-15 \sqrt {2}\, \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+52 \left (\cos ^{3}\left (f x +e \right )\right )-70 \left (\cos ^{2}\left (f x +e \right )\right )+30 \cos \left (f x +e \right )\right ) a}{15 c^{3} f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/c^3/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(-15*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+30*cos(f*x+e)*sin(f
*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)
/cos(f*x+e)*2^(1/2))-15*2^(1/2)*sin(f*x+e)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x
+e)*2^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+52*cos(f*x+e)^3-70*cos(f*x+e)^2+30*cos(f*x+e))/sin(f*x+e)/(-
1+cos(f*x+e))^2*a

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integra
l(a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x))/c**3

________________________________________________________________________________________