3.85 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac {15 a^3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}}+\frac {5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}} \]

[Out]

-15/4*a^3*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*(a+a*sec(f*x+e
))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/2)+5/4*(a^3+a^3*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)+15/4*a
^3*tan(f*x+e)/c^2/f/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3957, 3956, 3795, 203} \[ -\frac {15 a^3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}}+\frac {5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(-15*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f) - (a*(a + a*
Sec[e + f*x])^2*Tan[e + f*x])/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(4*
c*f*(c - c*Sec[e + f*x])^(3/2)) + (15*a^3*Tan[e + f*x])/(4*c^2*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))^3}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac {(5 a) \int \frac {\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 c}\\ &=-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac {\left (15 a^2\right ) \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx}{8 c^2}\\ &=-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}}+\frac {\left (15 a^3\right ) \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{4 c^2}\\ &=-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}}-\frac {\left (15 a^3\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 )}{2 c^2 f}\\ &=-\frac {15 a^3 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 4.57, size = 263, normalized size = 1.51 \[ -\frac {a^3 e^{-\frac {1}{2} i (2 e+f x)} \tan ^5\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 \left (120 e^{\frac {i e}{2}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )+\left (25 \cos \left (\frac {3}{2} (e+f x)\right )-9 \cos \left (\frac {5}{2} (e+f x)\right )\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)} \left (\cos \left (e+\frac {f x}{2}\right )+i \sin \left (e+\frac {f x}{2}\right )\right )\right )}{32 c^2 f (\sec (e+f x)-1)^2 \sqrt {\sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

-1/32*(a^3*Sec[(e + f*x)/2]*(1 + Sec[e + f*x])^3*(120*E^((I/2)*e)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)
))]*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))])] + (25
*Cos[(3*(e + f*x))/2] - 9*Cos[(5*(e + f*x))/2])*Csc[(e + f*x)/2]^4*Sqrt[Sec[e + f*x]]*(Cos[e + (f*x)/2] + I*Si
n[e + (f*x)/2]))*Tan[(e + f*x)/2]^5)/(c^2*E^((I/2)*(2*e + f*x))*f*(-1 + Sec[e + f*x])^2*Sqrt[Sec[e + f*x]]*Sqr
t[c - c*Sec[e + f*x]])

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fricas [A]  time = 0.58, size = 441, normalized size = 2.53 \[ \left [-\frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/8*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*sqrt(-c)*log((2*sqrt(2)*(cos(f*x + e)^2 + co
s(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*(9*a^3*cos(f*x + e)^3 - 8*a^3*cos(f*x + e)^2 - 13*a^3*cos(f*x + e) +
4*a^3)*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)), 1/4*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*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*(9*a^3*cos(f*x + e)^3 - 8*a^3*
cos(f*x + e)^2 - 13*a^3*cos(f*x + e) + 4*a^3)*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))]

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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))^3/(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)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)2/f*2*a^3*(1/33554432*(-14680064*sqrt(2)*sqrt(c*tan((f*x+exp(1))/2)^2-c)*(c*tan((f*x+exp(1))/2)^2-c)-18874
368*sqrt(2)*c*sqrt(c*tan((f*x+exp(1))/2)^2-c))/c^2/(c*tan((f*x+exp(1))/2)^2)^2/sign(tan((f*x+exp(1))/2))/sign(
tan((f*x+exp(1))/2)^2-1)-15/8/sqrt(c)/c^2*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqrt(c))/sqrt(2)/sign(tan((f*x+
exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)-1/sqrt(2)/c^2/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))

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maple [A]  time = 1.88, size = 206, normalized size = 1.18 \[ -\frac {a^{3} \left (15 \left (\cos ^{2}\left (f x +e \right )\right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-30 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )-18 \left (\cos ^{2}\left (f x +e \right )\right )+15 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+34 \cos \left (f x +e \right )-8\right ) \sin \left (f x +e \right )}{4 f \cos \left (f x +e \right )^{3} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(5/2),x)

[Out]

-1/4*a^3/f*(15*cos(f*x+e)^2*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
)-30*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)-18*cos(f*x
+e)^2+15*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)+34*cos(f*x+e)-8)*
sin(f*x+e)/cos(f*x+e)^3/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)

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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(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^3}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)),x)

[Out]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**(5/2),x)

[Out]

a**3*(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)
*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x) + Integral(3*sec(e + f*x)**2/(c**2*sqrt(-c*sec(e + f*x) +
c)*sec(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) + Int
egral(3*sec(e + f*x)**3/(c**2*sqrt(-c*sec(e + f*x) + c)*sec(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) + Integral(sec(e + f*x)**4/(c**2*sqrt(-c*sec(e + f*x) + c)*sec
(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))

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