\(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [99]
Optimal result
Integrand size = 30, antiderivative size = 146 \[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}
\]
[Out]
-a^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2)-a^2*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)/(a+a
*sec(f*x+e))^(1/2)+a^2*ln(1-cos(f*x+e))*tan(f*x+e)/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
Rubi [A] (verified)
Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3993, 3992, 3996, 31}
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac {a^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}
\]
[In]
Int[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^(5/2),x]
[Out]
-((a^2*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2))) - (a^2*Tan[e + f*x])/(c*f*Sqrt[a
+ a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) + (a^2*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(c^2*f*Sqrt[a + a*Se
c[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 3992
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 3993
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[-4*a^2*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/c, Int[Sqr
t[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 3996
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dist
[(-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]
Rubi steps \begin{align*}
\text {integral}& = -\frac {a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c} \\ & = -\frac {a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2} \\ & = -\frac {a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = -\frac {a^2 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.57
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\log (\cos (e+f x))+\log (1-\sec (e+f x))+\frac {-2+\sec (e+f x)}{(-1+\sec (e+f x))^2}\right ) \tan (e+f x)}{c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}}
\]
[In]
Integrate[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(a^2*(Log[Cos[e + f*x]] + Log[1 - Sec[e + f*x]] + (-2 + Sec[e + f*x])/(-1 + Sec[e + f*x])^2)*Tan[e + f*x])/(c^
2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Maple [A] (verified)
Time = 2.03 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.63
| | |
| method | result | size |
| | |
| default |
\(\frac {\sqrt {2}\, a \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-4 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+4 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \csc \left (f x +e \right )}{8 f \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\) |
\(238\) |
| risch |
\(\frac {a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (6 i {\mathrm e}^{i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )} f x -8 i {\mathrm e}^{2 i \left (f x +e \right )}-2 \,{\mathrm e}^{4 i \left (f x +e \right )} e +4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x -2 i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 \,{\mathrm e}^{3 i \left (f x +e \right )} e -6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x -12 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +4 \,{\mathrm e}^{i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 i {\mathrm e}^{3 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+6 i {\mathrm e}^{3 i \left (f x +e \right )}-2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e -f x -2 e \right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\) |
\(350\) |
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[In]
int((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/8/f*2^(1/2)*a*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^2/(c*(1-cos(f
*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)*csc(f*x+e)^2)^(5/2)*(1-cos(f*x+e))*(8*ln(-cot(f*x+e)+csc(f*x+e))*(1
-cos(f*x+e))^4*csc(f*x+e)^4-4*ln((1-cos(f*x+e))^2*csc(f*x+e)^2+1)*(1-cos(f*x+e))^4*csc(f*x+e)^4+4*(1-cos(f*x+e
))^2*csc(f*x+e)^2-1)*csc(f*x+e)
Fricas [F]
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
integral(-(a*sec(f*x + e) + a)^(3/2)*sqrt(-c*sec(f*x + e) + c)/(c^3*sec(f*x + e)^3 - 3*c^3*sec(f*x + e)^2 + 3*
c^3*sec(f*x + e) - c^3), x)
Sympy [F]
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx
\]
[In]
integrate((a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(5/2),x)
[Out]
Integral((a*(sec(e + f*x) + 1))**(3/2)/(-c*(sec(e + f*x) - 1))**(5/2), x)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1786 vs. \(2 (134) = 268\).
Time = 0.53 (sec) , antiderivative size = 1786, normalized size of antiderivative = 12.23
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\text {Too large to display}
\]
[In]
integrate((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
-((f*x + e)*a*cos(4*f*x + 4*e)^2 + 36*(f*x + e)*a*cos(2*f*x + 2*e)^2 + 16*(f*x + e)*a*cos(3/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e)))^2 + 16*(f*x + e)*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + (f*x
+ e)*a*sin(4*f*x + 4*e)^2 + 36*(f*x + e)*a*sin(2*f*x + 2*e)^2 + 16*(f*x + e)*a*sin(3/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e)))^2 + 16*(f*x + e)*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*(f*x + e)
*a*cos(2*f*x + 2*e) + (f*x + e)*a - 2*(a*cos(4*f*x + 4*e)^2 + 36*a*cos(2*f*x + 2*e)^2 + 16*a*cos(3/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + a*sin(4*
f*x + 4*e)^2 + 12*a*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*a*sin(2*f*x + 2*e)^2 + 16*a*sin(3/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*(6*a*cos(2*f
*x + 2*e) + a)*cos(4*f*x + 4*e) + 12*a*cos(2*f*x + 2*e) - 8*(a*cos(4*f*x + 4*e) + 6*a*cos(2*f*x + 2*e) - 4*a*c
os(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + a)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
- 8*(a*cos(4*f*x + 4*e) + 6*a*cos(2*f*x + 2*e) + a)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(
a*sin(4*f*x + 4*e) + 6*a*sin(2*f*x + 2*e) - 4*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(a*sin(4*f*x + 4*e) + 6*a*sin(2*f*x + 2*e))*sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + a)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 1) + 2*(6*(f*x + e)*a*cos(2*f*x + 2*e) + (f*x + e)*a - 4*a*sin
(2*f*x + 2*e))*cos(4*f*x + 4*e) - 2*(4*(f*x + e)*a*cos(4*f*x + 4*e) + 24*(f*x + e)*a*cos(2*f*x + 2*e) - 16*(f*
x + e)*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(f*x + e)*a + 3*a*sin(4*f*x + 4*e) + 2*a*sin
(2*f*x + 2*e))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*(4*(f*x + e)*a*cos(4*f*x + 4*e) + 24*(
f*x + e)*a*cos(2*f*x + 2*e) + 4*(f*x + e)*a + 3*a*sin(4*f*x + 4*e) + 2*a*sin(2*f*x + 2*e))*cos(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(3*(f*x + e)*a*sin(2*f*x + 2*e) + 2*a*cos(2*f*x + 2*e))*sin(4*f*x + 4*e)
- 8*a*sin(2*f*x + 2*e) - 2*(4*(f*x + e)*a*sin(4*f*x + 4*e) + 24*(f*x + e)*a*sin(2*f*x + 2*e) - 16*(f*x + e)*a
*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 3*a*cos(4*f*x + 4*e) - 2*a*cos(2*f*x + 2*e) - 3*a)*sin
(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*(4*(f*x + e)*a*sin(4*f*x + 4*e) + 24*(f*x + e)*a*sin(2*f
*x + 2*e) - 3*a*cos(4*f*x + 4*e) - 2*a*cos(2*f*x + 2*e) - 3*a)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2
*e))))*sqrt(a)*sqrt(c)/((c^3*cos(4*f*x + 4*e)^2 + 36*c^3*cos(2*f*x + 2*e)^2 + 16*c^3*cos(3/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e)))^2 + 16*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + c^3*sin(4*f*x
+ 4*e)^2 + 12*c^3*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 36*c^3*sin(2*f*x + 2*e)^2 + 16*c^3*sin(3/2*arctan2(sin(2
*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 16*c^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*c^3*cos
(2*f*x + 2*e) + c^3 + 2*(6*c^3*cos(2*f*x + 2*e) + c^3)*cos(4*f*x + 4*e) - 8*(c^3*cos(4*f*x + 4*e) + 6*c^3*cos(
2*f*x + 2*e) - 4*c^3*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + c^3)*cos(3/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) - 8*(c^3*cos(4*f*x + 4*e) + 6*c^3*cos(2*f*x + 2*e) + c^3)*cos(1/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) - 8*(c^3*sin(4*f*x + 4*e) + 6*c^3*sin(2*f*x + 2*e) - 4*c^3*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 8*(c^3*sin(4*f*x + 4*e) + 6
*c^3*sin(2*f*x + 2*e))*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)
Giac [F]
\[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
sage0*x
Mupad [F(-1)]
Timed out. \[
\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x
\]
[In]
int((a + a/cos(e + f*x))^(3/2)/(c - c/cos(e + f*x))^(5/2),x)
[Out]
int((a + a/cos(e + f*x))^(3/2)/(c - c/cos(e + f*x))^(5/2), x)