3.70 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}+\frac {a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}} \]

[Out]

1/16*a*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)-1/2*a*tan(f*x+e)/f/(c-c
*sec(f*x+e))^(5/2)+1/8*a*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3957, 3796, 3795, 203} \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}+\frac {a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(8*Sqrt[2]*c^(5/2)*f) - (a*Tan[e + f*x])
/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (a*Tan[e + f*x])/(8*c*f*(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 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))}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac {a \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 c}\\ &=-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}-\frac {a \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{16 c^2}\\ &=-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}+\frac {a \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 )}{8 c^2 f}\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} c^{5/2} f}-\frac {a \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a \tan (e+f x)}{8 c f (c-c \sec (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.12, size = 309, normalized size = 2.73 \[ \frac {a \left (-\frac {i \sqrt {2} \left (-1+e^{i (e+f x)}\right )^5 \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{5/2}}+48 \sin \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin ^5\left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)-48 \cos \left (\frac {e}{2}\right ) \cos \left (\frac {f x}{2}\right ) \sin ^5\left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)+56 \cot \left (\frac {e}{2}\right ) \sin ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)-16 \cot \left (\frac {e}{2}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)-56 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)+16 \csc \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sec ^3(e+f x)\right )}{16 c^2 f (\sec (e+f x)-1)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a*(((-I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^5*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))
])])/(1 + E^((2*I)*(e + f*x)))^(5/2) + 16*Csc[e/2]*Sec[e + f*x]^3*Sin[(f*x)/2]*Sin[(e + f*x)/2] - 16*Cot[e/2]*
Sec[e + f*x]^3*Sin[(e + f*x)/2]^2 - 56*Csc[e/2]*Sec[e + f*x]^3*Sin[(f*x)/2]*Sin[(e + f*x)/2]^3 + 56*Cot[e/2]*S
ec[e + f*x]^3*Sin[(e + f*x)/2]^4 - 48*Cos[e/2]*Cos[(f*x)/2]*Sec[e + f*x]^3*Sin[(e + f*x)/2]^5 + 48*Sec[e + f*x
]^3*Sin[e/2]*Sin[(f*x)/2]*Sin[(e + f*x)/2]^5))/(16*c^2*f*(-1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 405, normalized size = 3.58 \[ \left [-\frac {\sqrt {2} {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt {-c} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} - {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, {\left (3 \, a \cos \left (f x + e\right )^{3} + 4 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{32 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, -\frac {\sqrt {2} {\left (a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (3 \, a \cos \left (f x + e\right )^{3} + 4 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/32*(sqrt(2)*(a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(-c)*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*(3*a*cos(f*x + e)^3 + 4*a*cos(f*x + e)^2 + a*cos(f*x + e))*sqrt((c*cos(f*x
+ e) - c)/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)), -1/16*(sqrt(2)*
(a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)*sqrt(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) - 2*(3*a*cos(f*x + e)^3 + 4*a*cos(f*x + e)^2 + a*cos(f*x + e))*sq
rt((c*cos(f*x + e) - c)/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(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^(5/2),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/2*(1/8*(a*c*sqrt(c*tan((f*x+exp(1))/2)^2-c)*(c*tan((f*x+ex
p(1))/2)^2-c)-a*c^2*sqrt(c*tan((f*x+exp(1))/2)^2-c))/(c*tan((f*x+exp(1))/2)^2)^2+1/8*a*sqrt(c)*atan(sqrt(c*tan
((f*x+exp(1))/2)^2-c)/sqrt(c)))/sqrt(2)/c^3/sign(tan((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)

________________________________________________________________________________________

maple [B]  time = 1.42, size = 308, normalized size = 2.73 \[ -\frac {a \left (-1+\cos \left (f x +e \right )\right )^{3} \left (\left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+4 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}-\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-\left (\cos ^{2}\left (f x +e \right )\right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )+3 \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}}+2 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+2 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )-\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-\arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{2 f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{5} \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

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))^(5/2),x)

[Out]

-1/2*a/f*(-1+cos(f*x+e))^3*(cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+4*cos(f*x+e)*(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(3/2)-cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(1+co
s(f*x+e)))^(1/2))+3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)+2*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+2*c
os(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)))^(1/2)-arctan(1/(-2*cos
(f*x+e)/(1+cos(f*x+e)))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)^5/(-2*cos(f*x+e)/(1+cos(f*x+e)
))^(5/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)*sec(f*x + e)/(-c*sec(f*x + e) + c)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+\frac {a}{\cos \left (e+f\,x\right )}}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)),x)

[Out]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c-c*sec(f*x+e))**(5/2),x)

[Out]

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

________________________________________________________________________________________