3.99 \(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=146 \[ \frac {a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac {a^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]
[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)
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Rubi [A] time = 0.28, antiderivative size = 146, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used =
{3908, 3907, 3911, 31} \[ \frac {a^2 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a^2 \tan (e+f x)}{c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac {a^2 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
[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 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 3908
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 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]
Rubi steps
\begin {align*} \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 \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 ) \operatorname {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*}
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Mathematica [C] time = 1.27, size = 153, normalized size = 1.05 \[ \frac {a \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (6 \log \left (1-e^{i (e+f x)}\right )+\left (-8 \log \left (1-e^{i (e+f x)}\right )+4 i f x-6\right ) \cos (e+f x)+\left (2 \log \left (1-e^{i (e+f x)}\right )-i f x\right ) \cos (2 (e+f x))-3 i f x+4\right )}{2 c^2 f (\cos (e+f x)-1)^2 \sqrt {c-c \sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^(5/2),x]
[Out]
(a*(4 - (3*I)*f*x + Cos[e + f*x]*(-6 + (4*I)*f*x - 8*Log[1 - E^(I*(e + f*x))]) + 6*Log[1 - E^(I*(e + f*x))] +
Cos[2*(e + f*x)]*((-I)*f*x + 2*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(2*c^2*
f*(-1 + Cos[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sec \left (f x + e\right ) + c}}{c^{3} \sec \left (f x + e\right )^{3} - 3 \, c^{3} \sec \left (f x + e\right )^{2} + 3 \, c^{3} \sec \left (f x + e\right ) - c^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [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((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out
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maple [A] time = 2.10, size = 227, normalized size = 1.55 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (8 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-5 \left (\cos ^{2}\left (f x +e \right )\right )-16 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )+8 \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 \cos \left (f x +e \right )+8 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+3\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, a}{4 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(5/2),x)
[Out]
-1/4/f*(-1+cos(f*x+e))*(8*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))-4*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2-5*c
os(f*x+e)^2-16*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)+8*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*cos(f*x+e)+8*ln(
-(-1+cos(f*x+e))/sin(f*x+e))-4*ln(2/(1+cos(f*x+e)))+3)*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)/cos(f*x+
e)^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*a
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maxima [B] time = 1.21, size = 1786, normalized size = 12.23 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a/cos(e + f*x))^(3/2)/(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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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