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

Optimal. Leaf size=117 \[ -\frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {5 a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac {a^2 \tan (e+f x)}{f (c-c \sec (e+f x))^{5/2}} \]

[Out]

-3/8*a^2*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)-a^2*tan(f*x+e)/f/(c-c
*sec(f*x+e))^(5/2)+5/4*a^2*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)

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Rubi [A]  time = 0.24, antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3957, 3795, 203} \[ -\frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {3 a^2 \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{2 f (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 2.63, size = 359, normalized size = 3.07 \[ -\frac {a^2 \csc \left (\frac {e}{2}\right ) e^{-\frac {1}{2} i (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right ) \sec ^3\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)} (\sec (e+f x)+1)^2 \left (3 \sin \left (\frac {e}{2}\right ) \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \sin ^4\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )+\frac {e^{-\frac {3 i e}{2}} \left (-1+e^{i e}\right ) \left (\cos \left (\frac {f x}{2}\right )+i \sin \left (\frac {f x}{2}\right )\right ) \left (-9 e^{i e} \sin \left (\frac {f x}{2}\right )+9 e^{2 i e} \sin \left (\frac {f x}{2}\right )-e^{3 i e} \sin \left (\frac {3 f x}{2}\right )-9 i e^{i e} \left (1+e^{i e}\right ) \cos \left (\frac {f x}{2}\right )+i \left (1+e^{3 i e}\right ) \cos \left (\frac {3 f x}{2}\right )+\sin \left (\frac {3 f x}{2}\right )\right )}{16 \sqrt {\sec (e+f x)}}\right )}{4 c^2 f (\sec (e+f x)-1)^2 \sqrt {c-c \sec (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*(a^2*Csc[e/2]*Sec[(e + f*x)/2]^3*Sqrt[Sec[e + f*x]]*(1 + Sec[e + f*x])^2*(((-1 + E^(I*e))*(Cos[(f*x)/2] +
I*Sin[(f*x)/2])*((-9*I)*E^(I*e)*(1 + E^(I*e))*Cos[(f*x)/2] + I*(1 + E^((3*I)*e))*Cos[(3*f*x)/2] - 9*E^(I*e)*S
in[(f*x)/2] + 9*E^((2*I)*e)*Sin[(f*x)/2] + Sin[(3*f*x)/2] - E^((3*I)*e)*Sin[(3*f*x)/2]))/(16*E^(((3*I)/2)*e)*S
qrt[Sec[e + f*x]]) + 3*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))])]*Sin[e/2]*Sin[(e + f*x)/2]^4)*Tan[(e + f*x)/2])/(
c^2*E^((I/2)*(e + f*x))*f*(-1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

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fricas [A]  time = 0.56, size = 429, normalized size = 3.67 \[ \left [-\frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 4 \, a^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 4 \, a^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*sqrt(2)*(a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*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*(a^2*cos(f*x + e)^3 - 4*a^2*cos(f*x + e)^2 - 5*a^2*cos(f*x + e))*sqr
t((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/
8*(3*sqrt(2)*(a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*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*(a^2*cos(f*x + e)^3 - 4*a^2*cos(f*x + e)^2
- 5*a^2*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))]

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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))^(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*a^2*(3/8*sqrt(c)*atan(sqrt(c*tan((f*x+exp(1))/2)^2-c)/sqr
t(c))+1/8*(3*c*sqrt(c*tan((f*x+exp(1))/2)^2-c)*(c*tan((f*x+exp(1))/2)^2-c)+5*c^2*sqrt(c*tan((f*x+exp(1))/2)^2-
c))/(c*tan((f*x+exp(1))/2)^2)^2)/sqrt(2)/c^3/sign(tan((f*x+exp(1))/2))/sign(tan((f*x+exp(1))/2)^2-1)

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maple [B]  time = 1.90, size = 230, normalized size = 1.97 \[ -\frac {a^{2} \left (-1+\cos \left (f x +e \right )\right )^{3} \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+3 \left (\cos ^{2}\left (f x +e \right )\right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )-4 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-6 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )-5 \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+3 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}}\right )\right )}{f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{5} \left (-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/f*(-1+cos(f*x+e))^3*(cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+3*cos(f*x+e)^2*arctan(1/(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(1/2))-4*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-6*cos(f*x+e)*arctan(1/(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(1/2))-5*(-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))^(5/2)/sin(f*x+e)^5/(-2*cos(f*x+e)/(1+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))^2/(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 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + a/cos(e + f*x))^2/(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 \[ a^{2} \left (\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {2 \sec ^{2}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx\right ) \]

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))**(5/2),x)

[Out]

a**2*(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(2*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(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))

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