3.2.63 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{c+d \sec (e+f x)} \, dx\) [163]

Optimal. Leaf size=203 \[ \frac {2 a^3 \tan (e+f x)}{d f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}-\frac {2 a^{7/2} (c-d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{c d^{3/2} \sqrt {c+d} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}} \]

[Out]

2*a^3*tan(f*x+e)/d/f/(a+a*sec(f*x+e))^(1/2)+2*a^(7/2)*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c/f/(
a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-2*a^(7/2)*(c-d)^2*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/
(c+d)^(1/2))*tan(f*x+e)/c/d^(3/2)/f/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4025, 186, 65, 212, 214} \begin {gather*} -\frac {2 a^{7/2} (c-d)^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{c d^{3/2} f \sqrt {c+d} \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^{7/2} \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{c f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^3 \tan (e+f x)}{d f \sqrt {a \sec (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*Tan[e + f*x])/(d*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan
[e + f*x])/(c*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*a^(7/2)*(c - d)^2*ArcTanh[(Sqrt[d]*Sqr
t[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(c*d^(3/2)*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*
Sqrt[a + a*Sec[e + f*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 186

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 212

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

Rule 214

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

Rule 4025

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 6.61, size = 343, normalized size = 1.69 \begin {gather*} \frac {\cos ^{\frac {3}{2}}(e+f x) (d+c \cos (e+f x)) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (1+\sec (e+f x)))^{5/2} \left (\frac {10 (c-d)^2 (c+3 d+2 c \cos (e+f x)) \csc \left (\frac {1}{2} (e+f x)\right ) \left (-\tanh ^{-1}\left (\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )+\sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}\right )}{d (c+d) \sqrt {\cos (e+f x)} \sqrt {-\frac {d (-1+\sec (e+f x))}{c+d}}}+\frac {20 (3 c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}-\frac {16 (c-d)^2 d (d+c \cos (e+f x)) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {2 d \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )}{c+d}\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right )}{(c+d)^3 \cos ^{\frac {5}{2}}(e+f x)}+10 c \left (\sqrt {2} \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right )-\frac {2 \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )\right )}{40 c^2 f (c+d \sec (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]^(3/2)*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^5*(a*(1 + Sec[e + f*x]))^(5/2)*((10*(c - d)^2*(c + 3
*d + 2*c*Cos[e + f*x])*Csc[(e + f*x)/2]*(-ArcTanh[Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]] + Sqrt[-((d*(-1 +
Sec[e + f*x]))/(c + d))]))/(d*(c + d)*Sqrt[Cos[e + f*x]]*Sqrt[-((d*(-1 + Sec[e + f*x]))/(c + d))]) + (20*(3*c
- d)*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x]] - (16*(c - d)^2*d*(d + c*Cos[e + f*x])*Hypergeometric2F1[2, 5/2, 7/2
, (-2*d*Sec[e + f*x]*Sin[(e + f*x)/2]^2)/(c + d)]*Sin[(e + f*x)/2]^3)/((c + d)^3*Cos[e + f*x]^(5/2)) + 10*c*(S
qrt[2]*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]] - (2*Sin[(e + f*x)/2])/Sqrt[Cos[e + f*x]])))/(40*c^2*f*(c + d*Sec[e +
f*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1489\) vs. \(2(173)=346\).
time = 4.52, size = 1490, normalized size = 7.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1490\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(2*((c+d)*(c-d))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*
2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-ln(2*(2^(1/2)*(d/(c-d))^(1/2)*(-2*co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin
(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin
(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c^2*sin(f*x+e)+2*ln(2*(2^
(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)
/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)-((c+d)*(c-d))^(1/
2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/
2)*c*d*sin(f*x+e)-ln(2*(2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)-2^(1/2)*(d/(
c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*co
s(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c-d))*2^(1/2)*(-2*cos(
f*x+e)/(cos(f*x+e)+1))^(1/2)*d^2*sin(f*x+e)+ln(2*(-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2
)*c*sin(f*x+e)+2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*c
os(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*
x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c^2*sin(f*x+e)-2*ln(2*(-2^(1/2)*(d/(c-d))^(1/2)*(-2*co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)+2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin
(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin
(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*d*sin(f*x+e)+ln(2*(-2^(
1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*sin(f*x+e)+2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/
(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)+((c+d)*(c-d))^(1/2)*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2
))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d))*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2
)*d^2*sin(f*x+e)+4*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c*cos(f*x+e)-4*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*c)*(
a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)/sin(f*x+e)*a^2/c/((c+d)*(c-d))^(1/2)/d/(d/(c-d))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^(5/2)/(d*sec(f*x + e) + c), x)

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Fricas [A]
time = 7.95, size = 1210, normalized size = 5.96 \begin {gather*} \left [\frac {2 \, a^{2} c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2} + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a}{c d + d^{2}}} \log \left (\frac {2 \, {\left (c d + d^{2}\right )} \sqrt {-\frac {a}{c d + d^{2}}} \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 ) + {\left (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right ) + {\left (a^{2} d \cos \left (f x + e\right ) + a^{2} d\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 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 ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{c d f \cos \left (f x + e\right ) + c d f}, \frac {2 \, a^{2} c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 2 \, {\left (a^{2} d \cos \left (f x + e\right ) + a^{2} d\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2} + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a}{c d + d^{2}}} \log \left (\frac {2 \, {\left (c d + d^{2}\right )} \sqrt {-\frac {a}{c d + d^{2}}} \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 ) + {\left (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right )}{c d f \cos \left (f x + e\right ) + c d f}, \frac {2 \, a^{2} c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2} + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a}{c d + d^{2}}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {a}{c d + d^{2}}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a \sin \left (f x + e\right )}\right ) + {\left (a^{2} d \cos \left (f x + e\right ) + a^{2} d\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} - 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 ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{c d f \cos \left (f x + e\right ) + c d f}, \frac {2 \, {\left (a^{2} c \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2} + {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a}{c d + d^{2}}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {a}{c d + d^{2}}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{a \sin \left (f x + e\right )}\right ) - {\left (a^{2} d \cos \left (f x + e\right ) + a^{2} d\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right )\right )}}{c d f \cos \left (f x + e\right ) + c d f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2
*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(-a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x
+ e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + (a^2*d*cos(f*x + e) + a^2*d)*sqrt(-a)*log((2*a*cos(f*x
+ e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(c
os(f*x + e) + 1)))/(c*d*f*cos(f*x + e) + c*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)
- 2*(a^2*d*cos(f*x + e) + a^2*d)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)
*sin(f*x + e))) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(-a/(c*d
+ d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x +
e) + (a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e)
+ d)))/(c*d*f*cos(f*x + e) + c*d*f), (2*a^2*c*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + 2*(a^2*c^
2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqr
t(a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(a*sin(f*x + e))) + (a^2*d*cos(f*x + e)
+ a^2*d)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*si
n(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/(c*d*f*cos(f*x + e) + c*d*f), 2*(a^2*c*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sin(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x
+ e))*sqrt(a/(c*d + d^2))*arctan((c + d)*sqrt(a/(c*d + d^2))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x
+ e)/(a*sin(f*x + e))) - (a^2*d*cos(f*x + e) + a^2*d)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c
os(f*x + e)/(sqrt(a)*sin(f*x + e))))/(c*d*f*cos(f*x + e) + c*d*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}{c + d \sec {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(e + f*x) + 1))**(5/2)/(c + d*sec(e + f*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (173) = 346\).
time = 2.03, size = 412, normalized size = 2.03 \begin {gather*} -\frac {\frac {2 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} a^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )} d} + \frac {\sqrt {-a} a^{2} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{c} - \frac {\sqrt {-a} a^{2} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{c} + \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-a} a^{3} c^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 2 \, \sqrt {2} \sqrt {-a} a^{3} c d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + \sqrt {2} \sqrt {-a} a^{3} d^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} c - {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} d + a c + 3 \, a d\right )}}{4 \, \sqrt {-c d - d^{2}} a}\right )}{\sqrt {-c d - d^{2}} a c d}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*sqrt(2)*sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a)*a^3*sgn(cos(f*x + e))*tan(1/2*f*x + 1/2*e)/((a*tan(1/2*f*x + 1
/2*e)^2 - a)*d) + sqrt(-a)*a^2*log(abs((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2
- a*(2*sqrt(2) + 3)))*sgn(cos(f*x + e))/c - sqrt(-a)*a^2*log(abs((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan
(1/2*f*x + 1/2*e)^2 + a))^2 + a*(2*sqrt(2) - 3)))*sgn(cos(f*x + e))/c + sqrt(2)*(sqrt(2)*sqrt(-a)*a^3*c^2*sgn(
cos(f*x + e)) - 2*sqrt(2)*sqrt(-a)*a^3*c*d*sgn(cos(f*x + e)) + sqrt(2)*sqrt(-a)*a^3*d^2*sgn(cos(f*x + e)))*arc
tan(1/4*sqrt(2)*((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*c - (sqrt(-a)*tan(1/2
*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*d + a*c + 3*a*d)/(sqrt(-c*d - d^2)*a))/(sqrt(-c*d - d^2
)*a*c*d))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{c+\frac {d}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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