\(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx\) [62]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}+\frac {8 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 c^3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3989, 3972, 472, 209} \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^3 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^3 f}+\frac {8 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 c^3 f} \]

[In]

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

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^3*f) + (2*a^2*Cot[e + f*x]*Sqrt[a + a*S
ec[e + f*x]])/(c^3*f) + (8*Cot[e + f*x]^5*(a + a*Sec[e + f*x])^(5/2))/(5*c^3*f)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3972

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 3989

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*} \text {integral}& = -\frac {\int \cot ^6(e+f x) (a+a \sec (e+f x))^{11/2} \, dx}{a^3 c^3} \\ & = \frac {2 \text {Subst}\left (\int \frac {\left (2+a x^2\right )^2}{x^6 \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 \text {Subst}\left (\int \left (\frac {4}{x^6}+\frac {a^2}{x^2}-\frac {a^3}{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^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}+\frac {8 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 c^3 f}-\frac {\left (2 a^3\right ) \text {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^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}+\frac {8 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 c^3 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^3 \left (4+3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},1-\sec (e+f x)\right )+5 \sec (e+f x)\right ) \tan (e+f x)}{15 c^3 f (-1+\sec (e+f x))^3 \sqrt {a (1+\sec (e+f x))}} \]

[In]

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

[Out]

(2*a^3*(4 + 3*Hypergeometric2F1[-5/2, 1, -3/2, 1 - Sec[e + f*x]] + 5*Sec[e + f*x])*Tan[e + f*x])/(15*c^3*f*(-1
+ Sec[e + f*x])^3*Sqrt[a*(1 + Sec[e + f*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(92)=184\).

Time = 70.08 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.57

method	result	size
default	\(\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (5 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \cos \left (f x +e \right )^{2}-10 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+9 \cos \left (f x +e \right )^{2} \cot \left (f x +e \right )+5 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-10 \cos \left (f x +e \right ) \cot \left (f x +e \right )+5 \cot \left (f x +e \right )\right )}{5 c^{3} f \left (\cos \left (f x +e \right )-1\right )^{2}}\)	\(267\)

[In]

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

[Out]

2/5/c^3/f*a^2*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)-1)^2*(5*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctanh(sin(f*x+
e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*cos(f*x+e)^2-10*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(
f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+9*cos(f*x+e)^2*cot(f*x+e)+5*arctan
h(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-10*cos(f*x+
e)*cot(f*x+e)+5*cot(f*x+e))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (92) = 184\).

Time = 0.31 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.24 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\left [\frac {5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (9 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{10 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (9 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[1/10*(5*(a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*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*(9*a^2*cos(f*x + e)^3 - 10*a^2*cos(f*x + e)^2 + 5*a^2*cos(f*x + e))*sqrt((a*co
s(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)), 1/5*(5*(a
^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*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*(9*a^2*cos(f*x + e)^3 -
10*a^2*cos(f*x + e)^2 + 5*a^2*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))]

Sympy [F]

\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=- \frac {\int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx}{c^{3}} \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (c \sec \left (f x + e\right ) - c\right )}^{3}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]

[In]

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

[Out]

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