3.2.55 \(\int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2 \, dx\) [155]

Optimal. Leaf size=176 \[ \frac {2 a^{5/2} c^2 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}} \]

[Out]

2/5*a^2*(c+d*sec(f*x+e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2/15*a^2*(12*c^2+50*c*d+18*d^2+d*(4*c+9*d)*sec(
f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)+2*a^(5/2)*c^2*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/f
/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4025, 158, 152, 65, 212} \begin {gather*} \frac {2 a^{5/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \tan (e+f x) \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right )}{15 f \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \tan (e+f x) (c+d \sec (e+f x))^2}{5 f \sqrt {a \sec (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*c^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*
Sec[e + f*x]]) + (2*a^2*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(2*(6*c^2
+ 25*c*d + 9*d^2) + d*(4*c + 9*d)*Sec[e + f*x])*Tan[e + f*x])/(15*f*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 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 158

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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 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 (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2 \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x) (c+d x)^2}{x \sqrt {a-a x}} \, 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^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}+\frac {(2 a \tan (e+f x)) \text {Subst}\left (\int \frac {(c+d x) \left (-\frac {5 a^2 c}{2}-\frac {1}{2} a^2 (4 c+9 d) x\right )}{x \sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{5 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}-\frac {\left (a^3 c^2 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-a x}} \, 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^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}+\frac {\left (2 a^2 c^2 \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 )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {2 a^{5/2} c^2 \tanh ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt {a+a \sec (e+f x)}}+\frac {2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.38, size = 145, normalized size = 0.82 \begin {gather*} \frac {a \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt {a (1+\sec (e+f x))} \left (30 \sqrt {2} c^2 \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \cos ^{\frac {5}{2}}(e+f x)+2 \left (15 c^2+50 c d+24 d^2+2 d (10 c+9 d) \cos (e+f x)+\left (15 c^2+50 c d+18 d^2\right ) \cos (2 (e+f x))\right ) \sin \left (\frac {1}{2} (e+f x)\right )\right )}{30 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*(30*Sqrt[2]*c^2*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]]
*Cos[e + f*x]^(5/2) + 2*(15*c^2 + 50*c*d + 24*d^2 + 2*d*(10*c + 9*d)*Cos[e + f*x] + (15*c^2 + 50*c*d + 18*d^2)
*Cos[2*(e + f*x)])*Sin[(e + f*x)/2]))/(30*f)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(381\) vs. \(2(155)=310\).
time = 2.81, size = 382, normalized size = 2.17

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (15 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}\, \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} c^{2}+30 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} c^{2}+15 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) c^{2} \sin \left (f x +e \right )+120 \left (\cos ^{3}\left (f x +e \right )\right ) c^{2}+400 \left (\cos ^{3}\left (f x +e \right )\right ) c d +144 \left (\cos ^{3}\left (f x +e \right )\right ) d^{2}-120 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2}-320 \left (\cos ^{2}\left (f x +e \right )\right ) c d -72 \left (\cos ^{2}\left (f x +e \right )\right ) d^{2}-80 \cos \left (f x +e \right ) c d -48 \cos \left (f x +e \right ) d^{2}-24 d^{2}\right ) a}{60 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )}\)	\(382\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(15*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/co
s(f*x+e)*2^(1/2))*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*c^2+30*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))*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(
cos(f*x+e)+1))^(5/2)*c^2+15*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*2^(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))*c^2*sin(f*x+e)+120*cos(f*x+e)^3*c^2+400*cos(f*x+e)^3*c*d+144*cos(f*
x+e)^3*d^2-120*cos(f*x+e)^2*c^2-320*cos(f*x+e)^2*c*d-72*cos(f*x+e)^2*d^2-80*cos(f*x+e)*c*d-48*cos(f*x+e)*d^2-2
4*d^2)/cos(f*x+e)^2/sin(f*x+e)*a

________________________________________________________________________________________

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))^(3/2)*(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/30*(15*((a*c^2*cos(2*f*x + 2*e)^2 + a*c^2*sin(2*f*x + 2*e)^2 + 2*a*c^2*cos(2*f*x + 2*e) + a*c^2)*arctan2((c
os(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1) - (a*c^2*cos(2*f*x + 2*e)^2 + a*c^2*sin(2*f*x + 2*e)^2 + 2*a*c^2*c
os(2*f*x + 2*e) + a*c^2)*arctan2((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*sin(
1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)), (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x +
2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 1) - 2*(a*c^2*f*cos(2*f*x + 2*e)^2
+ a*c^2*f*sin(2*f*x + 2*e)^2 + 2*a*c^2*f*cos(2*f*x + 2*e) + a*c^2*f)*integrate((cos(2*f*x + 2*e)^2 + sin(2*f*
x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(((cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x
+ 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2
*f*x + 2*e)^2)*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*c
os(2*f*x + 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*si
n(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) -
((cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e)
- 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (cos(6*f*x + 6*
e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x +
2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e)^2)*sin(7/2*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) + 1)))/((cos(2*f*x + 2*e)^4 + sin(2*f*x + 2*e)^4 +
(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e)^2 + 4*(cos(2*f*x + 2*e)^2
+ sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e)^2 + 2*cos(2*f*x + 2*e)^3 + (cos(2*f*x + 2*e)^
2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(6*f*x + 6*e)^2 + 4*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)
^2 + 2*cos(2*f*x + 2*e) + 1)*sin(4*f*x + 4*e)^2 + (2*cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x +
2*e)^2 + 2*(cos(2*f*x + 2*e)^3 + cos(2*f*x + 2*e)*sin(2*f*x + 2*e)^2 + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e
)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 2*cos(2*f*x + 2*e)^2 + cos(2*f*x + 2*e))*cos(6*f*x + 6*e) + 4
*(cos(2*f*x + 2*e)^3 + cos(2*f*x + 2*e)*sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e)^2 + cos(2*f*x + 2*e))*cos(4*f*
x + 4*e) + cos(2*f*x + 2*e)^2 + 2*(sin(2*f*x + 2*e)^3 + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f
*x + 2*e) + 1)*sin(4*f*x + 4*e) + (cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x + 2*e))*sin(6*f*x +
6*e) + 4*(sin(2*f*x + 2*e)^3 + (cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x + 2*e))*sin(4*f*x + 4*e
))*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + (cos(2*f*x + 2*e)^4 + sin(2*f*x + 2*e)^4 + (co
s(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e)^2 + 4*(cos(2*f*x + 2*e)^2 + s
in(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e)^2 + 2*cos(2*f*x + 2*e)^3 + (cos(2*f*x + 2*e)^2 +
sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(6*f*x + 6*e)^2 + 4*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)*sin(4*f*x + 4*e)^2 + (2*cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x + 2*e)
^2 + 2*(cos(2*f*x + 2*e)^3 + cos(2*f*x + 2*e)*sin(2*f*x + 2*e)^2 + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2
+ 2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 2*cos(2*f*x + 2*e)^2 + cos(2*f*x + 2*e))*cos(6*f*x + 6*e) + 4*(co
s(2*f*x + 2*e)^3 + cos(2*f*x + 2*e)*sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e)^2 + cos(2*f*x + 2*e))*cos(4*f*x +
4*e) + cos(2*f*x + 2*e)^2 + 2*(sin(2*f*x + 2*e)^3 + 2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x +
2*e) + 1)*sin(4*f*x + 4*e) + (cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x + 2*e))*sin(6*f*x + 6*e)
+ 4*(sin(2*f*x + 2*e)^3 + (cos(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*sin(2*f*x + 2*e))*sin(4*f*x + 4*e))*s
in(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2), x) - 2*((5*a*c^2 + 12*a*c*d + 4*a*d^2)*f*cos(2*f*x
+ 2*e)^2 + (5*a*c^2 + 12*a*c*d + 4*a*d^2)*f*sin(2*f*x + 2*e)^2 + 2*(5*a*c^2 + 12*a*c*d + 4*a*d^2)*f*cos(2*f*x
+ 2*e) + (5*a*c^2 + 12*a*c*d + 4*a*d^2)*f)*integrate((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x +
2*e) + 1)^(1/4)*(((cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)
^2 + sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e)^2)*cos(5/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x
+ 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*...

________________________________________________________________________________________

Fricas [A]
time = 2.99, size = 427, normalized size = 2.43 \begin {gather*} \left [\frac {15 \, {\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\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 ) + 2 \, {\left (3 \, a d^{2} + {\left (15 \, a c^{2} + 50 \, a c d + 18 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (10 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\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 (3 \, a d^{2} + {\left (15 \, a c^{2} + 50 \, a c d + 18 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left (10 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{15 \, {\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(a*c^2*cos(f*x + e)^3 + a*c^2*cos(f*x + e)^2)*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)/(cos(f*x + e) + 1)) + 2*(3*a*d
^2 + (15*a*c^2 + 50*a*c*d + 18*a*d^2)*cos(f*x + e)^2 + (10*a*c*d + 9*a*d^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)
+ a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^2), -2/15*(15*(a*c^2*cos(f*x + e)^3 + a*c
^2*cos(f*x + e)^2)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))
- (3*a*d^2 + (15*a*c^2 + 50*a*c*d + 18*a*d^2)*cos(f*x + e)^2 + (10*a*c*d + 9*a*d^2)*cos(f*x + e))*sqrt((a*cos
(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^2)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (c + d \sec {\left (e + f x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sec(e + f*x) + 1))**(3/2)*(c + d*sec(e + f*x))**2, x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (155) = 310\).
time = 1.56, size = 359, normalized size = 2.04 \begin {gather*} -\frac {\frac {15 \, \sqrt {-a} a^{2} c^{2} \log \left (\frac {{\left | 2 \, {\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} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\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} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left ({\left (\sqrt {2} {\left (15 \, a^{4} c^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 40 \, a^{4} c d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 12 \, a^{4} d^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, \sqrt {2} {\left (3 \, a^{4} c^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 10 \, a^{4} c d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 3 \, a^{4} d^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, \sqrt {2} {\left (a^{4} c^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 4 \, a^{4} c d \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{4} d^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\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 )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*sqrt(-a)*a^2*c^2*log(abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 -
4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 4*sqr
t(2)*abs(a) - 6*a))*sgn(cos(f*x + e))/abs(a) - 2*((sqrt(2)*(15*a^4*c^2*sgn(cos(f*x + e)) + 40*a^4*c*d*sgn(cos(
f*x + e)) + 12*a^4*d^2*sgn(cos(f*x + e)))*tan(1/2*f*x + 1/2*e)^2 - 10*sqrt(2)*(3*a^4*c^2*sgn(cos(f*x + e)) + 1
0*a^4*c*d*sgn(cos(f*x + e)) + 3*a^4*d^2*sgn(cos(f*x + e))))*tan(1/2*f*x + 1/2*e)^2 + 15*sqrt(2)*(a^4*c^2*sgn(c
os(f*x + e)) + 4*a^4*c*d*sgn(cos(f*x + e)) + 2*a^4*d^2*sgn(cos(f*x + e))))*tan(1/2*f*x + 1/2*e)/((a*tan(1/2*f*
x + 1/2*e)^2 - a)^2*sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a)))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________