3.109 \(\int \frac{(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=244 \[ -\frac{a^3 \tan (e+f x)}{c^4 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{7/2}}+\frac{a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^5 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{11/2}} \]

[Out]

(-4*a^3*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(11/2)) - (a^3*Tan[e + f*x])/(3*c^2*f
*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(7/2)) - (a^3*Tan[e + f*x])/(2*c^3*f*Sqrt[a + a*Sec[e + f*x]]*(
c - c*Sec[e + f*x])^(5/2)) - (a^3*Tan[e + f*x])/(c^4*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) +
(a^3*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(c^5*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.47412, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3910, 3907, 3911, 31} \[ -\frac{a^3 \tan (e+f x)}{c^4 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{7/2}}+\frac{a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^5 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(11/2),x]

[Out]

(-4*a^3*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(11/2)) - (a^3*Tan[e + f*x])/(3*c^2*f
*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(7/2)) - (a^3*Tan[e + f*x])/(2*c^3*f*Sqrt[a + a*Sec[e + f*x]]*(
c - c*Sec[e + f*x])^(5/2)) - (a^3*Tan[e + f*x])/(c^4*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) +
(a^3*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(c^5*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3910

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-8*a^3*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*
d, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)]

Rule 3907

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[
(-2*a*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/c, Int[Sqrt[a +
b*Csc[e + f*x]]*(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] && LtQ[n, -2^(-1)]

Rule 3911

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{11/2}} \, dx &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{7/2}} \, dx}{c^2}\\ &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx}{c^3}\\ &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c^4}\\ &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{a^3 \tan (e+f x)}{c^4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{a^2 \int \frac{\sqrt{a+a \sec (e+f x)}}{\sqrt{c-c \sec (e+f x)}} \, dx}{c^5}\\ &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{a^3 \tan (e+f x)}{c^4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{\left (a^3 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c^4 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{4 a^3 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{11/2}}-\frac{a^3 \tan (e+f x)}{3 c^2 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac{a^3 \tan (e+f x)}{2 c^3 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac{a^3 \tan (e+f x)}{c^4 f \sqrt{a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac{a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^5 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 5.94935, size = 299, normalized size = 1.23 \[ \frac{\sin ^{11}\left (\frac{1}{2} (e+f x)\right ) \sec ^{\frac{11}{2}}(e+f x) (a (\sec (e+f x)+1))^{5/2} \left (-\frac{(5612 \cos (e+f x)-5 (736 \cos (2 (e+f x))-367 \cos (3 (e+f x))+111 \cos (4 (e+f x))-21 \cos (5 (e+f x))+625)) \csc ^{10}\left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)} \sqrt{\sec (e+f x)+1}}{240 f}+\frac{32 i \sqrt{2} e^{\frac{1}{2} i (e+f x)} \sqrt{\frac{\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}} \left (f x+2 i \log \left (1-e^{i (e+f x)}\right )\right )}{f \left (1+e^{i (e+f x)}\right ) \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}}}\right )}{(\sec (e+f x)+1)^{5/2} (c-c \sec (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(11/2),x]

[Out]

(Sec[e + f*x]^(11/2)*(a*(1 + Sec[e + f*x]))^(5/2)*(((32*I)*Sqrt[2]*E^((I/2)*(e + f*x))*Sqrt[(1 + E^(I*(e + f*x
)))^2/(1 + E^((2*I)*(e + f*x)))]*(f*x + (2*I)*Log[1 - E^(I*(e + f*x))]))/((1 + E^(I*(e + f*x)))*Sqrt[E^(I*(e +
f*x))/(1 + E^((2*I)*(e + f*x)))]*f) - ((5612*Cos[e + f*x] - 5*(625 + 736*Cos[2*(e + f*x)] - 367*Cos[3*(e + f*
x)] + 111*Cos[4*(e + f*x)] - 21*Cos[5*(e + f*x)]))*Csc[(e + f*x)/2]^10*Sec[(e + f*x)/2]*Sqrt[Sec[e + f*x]]*Sqr
t[1 + Sec[e + f*x]])/(240*f))*Sin[(e + f*x)/2]^11)/((1 + Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(11/2))

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Maple [A]  time = 0.292, size = 415, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(11/2),x)

[Out]

-1/120/f*a^2*(-1+cos(f*x+e))*(240*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^5-120*ln(2/(1+cos(f*x+e)))*cos(f*
x+e)^5-233*cos(f*x+e)^5-1200*cos(f*x+e)^4*ln(-(-1+cos(f*x+e))/sin(f*x+e))+600*cos(f*x+e)^4*ln(2/(1+cos(f*x+e))
)+325*cos(f*x+e)^4+2400*cos(f*x+e)^3*ln(-(-1+cos(f*x+e))/sin(f*x+e))-1200*cos(f*x+e)^3*ln(2/(1+cos(f*x+e)))-11
0*cos(f*x+e)^3-2400*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+1200*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2-290*co
s(f*x+e)^2+1200*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-600*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+295*cos(f*x+e)-
240*ln(-(-1+cos(f*x+e))/sin(f*x+e))+120*ln(2/(1+cos(f*x+e)))-83)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/(c*(-1+
cos(f*x+e))/cos(f*x+e))^(11/2)/sin(f*x+e)/cos(f*x+e)^5

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\right )} \sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{c^{6} \sec \left (f x + e\right )^{6} - 6 \, c^{6} \sec \left (f x + e\right )^{5} + 15 \, c^{6} \sec \left (f x + e\right )^{4} - 20 \, c^{6} \sec \left (f x + e\right )^{3} + 15 \, c^{6} \sec \left (f x + e\right )^{2} - 6 \, c^{6} \sec \left (f x + e\right ) + c^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

integral((a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2)*sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(c
^6*sec(f*x + e)^6 - 6*c^6*sec(f*x + e)^5 + 15*c^6*sec(f*x + e)^4 - 20*c^6*sec(f*x + e)^3 + 15*c^6*sec(f*x + e)
^2 - 6*c^6*sec(f*x + e) + c^6), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(11/2),x, algorithm="giac")

[Out]

Timed out