3.3.2 \(\int (a+b \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx\) [202]
Optimal. Leaf size=442 \[ -\frac {2 (a-b) \sqrt {a+b} \left (35 a b c+23 a^2 d+9 b^2 d\right ) \cot (e+f x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{15 b f}+\frac {2 \sqrt {a+b} \left (a^2 b (45 c-23 d)-a b^2 (35 c-17 d)+b^3 (5 c-9 d)+15 a^3 d\right ) \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{15 b f}-\frac {2 a^2 \sqrt {a+b} c \cot (e+f x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{f}+\frac {2 b (5 b c+8 a d) \sqrt {a+b \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f} \]
[Out]
-2/15*(a-b)*(23*a^2*d+35*a*b*c+9*b^2*d)*cot(f*x+e)*EllipticE((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^
(1/2))*(a+b)^(1/2)*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/b/f+2/15*(a^2*b*(45*c-23*d)-
a*b^2*(35*c-17*d)+b^3*(5*c-9*d)+15*a^3*d)*cot(f*x+e)*EllipticF((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b)
)^(1/2))*(a+b)^(1/2)*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/b/f-2*a^2*c*cot(f*x+e)*Ell
ipticPi((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(f*x+e))/(a+b))^(
1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/f+2/5*b*d*(a+b*sec(f*x+e))^(3/2)*tan(f*x+e)/f+2/15*b*(8*a*d+5*b*c)*(a+b*s
ec(f*x+e))^(1/2)*tan(f*x+e)/f
________________________________________________________________________________________
Rubi [A]
time = 0.43, antiderivative size = 442, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4003, 4141,
4143, 4006, 3869, 3917, 4089} \begin {gather*} -\frac {2 (a-b) \sqrt {a+b} \left (23 a^2 d+35 a b c+9 b^2 d\right ) \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b f}-\frac {2 a^2 c \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{f}+\frac {2 \sqrt {a+b} \left (15 a^3 d+a^2 b (45 c-23 d)-a b^2 (35 c-17 d)+b^3 (5 c-9 d)\right ) \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b f}+\frac {2 b (8 a d+5 b c) \tan (e+f x) \sqrt {a+b \sec (e+f x)}}{15 f}+\frac {2 b d \tan (e+f x) (a+b \sec (e+f x))^{3/2}}{5 f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x]),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*(35*a*b*c + 23*a^2*d + 9*b^2*d)*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/
(15*b*f) + (2*Sqrt[a + b]*(a^2*b*(45*c - 23*d) - a*b^2*(35*c - 17*d) + b^3*(5*c - 9*d) + 15*a^3*d)*Cot[e + f*x
]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)
]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(15*b*f) - (2*a^2*Sqrt[a + b]*c*Cot[e + f*x]*EllipticPi[(a + b)/a,
ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*
(1 + Sec[e + f*x]))/(a - b))])/f + (2*b*(5*b*c + 8*a*d)*Sqrt[a + b*Sec[e + f*x]]*Tan[e + f*x])/(15*f) + (2*b*d
*(a + b*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*f)
Rule 3869
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b
*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b
*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Rule 3917
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4003
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-b)
*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2
*c*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x],
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
Rule 4006
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 4089
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rule 4141
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), I
nt[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) +
a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Rule 4143
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[Csc[e + f*x
]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]
Rubi steps
\begin {align*} \int (a+b \sec (e+f x))^{5/2} (c+d \sec (e+f x)) \, dx &=\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {2}{5} \int \sqrt {a+b \sec (e+f x)} \left (\frac {5 a^2 c}{2}+\frac {1}{2} \left (10 a b c+5 a^2 d+3 b^2 d\right ) \sec (e+f x)+\frac {1}{2} b (5 b c+8 a d) \sec ^2(e+f x)\right ) \, dx\\ &=\frac {2 b (5 b c+8 a d) \sqrt {a+b \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {4}{15} \int \frac {\frac {15 a^3 c}{4}+\frac {1}{4} \left (45 a^2 b c+5 b^3 c+15 a^3 d+17 a b^2 d\right ) \sec (e+f x)+\frac {1}{4} b \left (35 a b c+23 a^2 d+9 b^2 d\right ) \sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx\\ &=\frac {2 b (5 b c+8 a d) \sqrt {a+b \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\frac {4}{15} \int \frac {\frac {15 a^3 c}{4}+\left (-\frac {1}{4} b \left (35 a b c+23 a^2 d+9 b^2 d\right )+\frac {1}{4} \left (45 a^2 b c+5 b^3 c+15 a^3 d+17 a b^2 d\right )\right ) \sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx+\frac {1}{15} \left (b \left (35 a b c+23 a^2 d+9 b^2 d\right )\right ) \int \frac {\sec (e+f x) (1+\sec (e+f x))}{\sqrt {a+b \sec (e+f x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b c+23 a^2 d+9 b^2 d\right ) \cot (e+f x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{15 b f}+\frac {2 b (5 b c+8 a d) \sqrt {a+b \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}+\left (a^3 c\right ) \int \frac {1}{\sqrt {a+b \sec (e+f x)}} \, dx+\frac {1}{15} \left (a^2 b (45 c-23 d)-a b^2 (35 c-17 d)+b^3 (5 c-9 d)+15 a^3 d\right ) \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b c+23 a^2 d+9 b^2 d\right ) \cot (e+f x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{15 b f}+\frac {2 \sqrt {a+b} \left (a^2 b (45 c-23 d)-a b^2 (35 c-17 d)+b^3 (5 c-9 d)+15 a^3 d\right ) \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{15 b f}-\frac {2 a^2 \sqrt {a+b} c \cot (e+f x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{f}+\frac {2 b (5 b c+8 a d) \sqrt {a+b \sec (e+f x)} \tan (e+f x)}{15 f}+\frac {2 b d (a+b \sec (e+f x))^{3/2} \tan (e+f x)}{5 f}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(7138\) vs. \(2(442)=884\).
time = 25.49, size = 7138, normalized size = 16.15 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sec[e + f*x])^(5/2)*(c + d*Sec[e + f*x]),x]
[Out]
Result too large to show
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3284\) vs.
\(2(403)=806\).
time = 3.45, size = 3285, normalized size = 7.43
| | |
| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(3285\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x,method=_RETURNVERBOSE)
[Out]
-2/15/f*(cos(f*x+e)+1)^2*((a*cos(f*x+e)+b)/cos(f*x+e))^(1/2)*(-1+cos(f*x+e))^2*(23*cos(f*x+e)^4*a^3*d-23*cos(f
*x+e)^3*a^3*d+9*cos(f*x+e)^3*b^3*d-6*cos(f*x+e)^2*b^3*d+5*cos(f*x+e)^3*b^3*c-5*cos(f*x+e)*b^3*c+35*cos(f*x+e)^
4*a^2*b*c+11*cos(f*x+e)^4*a^2*b*d+5*cos(f*x+e)^4*a*b^2*c+9*cos(f*x+e)^4*a*b^2*d-35*cos(f*x+e)^3*a^2*b*c+23*cos
(f*x+e)^3*a^2*b*d+35*cos(f*x+e)^3*a*b^2*c+5*cos(f*x+e)^3*a*b^2*d-34*cos(f*x+e)^2*a^2*b*d-40*cos(f*x+e)^2*a*b^2
*c-14*cos(f*x+e)*a*b^2*d-15*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f
*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^3*c+15*cos(f*x+e)^3*sin(f*x+
e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/s
in(f*x+e),((a-b)/(a+b))^(1/2))*a^3*d+5*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e
)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3*c+9*cos(f*x+e)^
3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos
(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3*d-23*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((
a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^3*d-9*
cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*Ellipt
icE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3*d+30*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1
))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^
(1/2))*a^3*c-15*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+
b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^3*c+15*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+
e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((
a-b)/(a+b))^(1/2))*a^3*d+5*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*
x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3*c+9*cos(f*x+e)^2*sin(f*x+e)
*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin
(f*x+e),((a-b)/(a+b))^(1/2))*b^3*d-23*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)
+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^3*d-9*cos(f*x+e)^2
*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(
f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^3*d+30*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a
*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^3*c
-3*b^3*d+45*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^
(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b*c+23*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)
/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-
b)/(a+b))^(1/2))*a^2*b*d+35*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f
*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2*c+17*cos(f*x+e)^3*sin(f*
x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))
/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2*d-35*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(
f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b*c-23*cos
(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE
((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b*d-35*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1)
)^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2)
)*a*b^2*c-9*cos(f*x+e)^3*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^
(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2*d+45*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)
/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-
b)/(a+b))^(1/2))*a^2*b*c+23*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f
*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*b*d+35*cos(f*x+e)^2*sin(f*
x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))
/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2*c+17*cos(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(
f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b^2*d-35*cos
(f*x+e)^2*sin(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1)...
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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+b*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e) + a)^(5/2)*(d*sec(f*x + e) + c), x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="fricas")
[Out]
integral((b^2*d*sec(f*x + e)^3 + a^2*c + (b^2*c + 2*a*b*d)*sec(f*x + e)^2 + (2*a*b*c + a^2*d)*sec(f*x + e))*sq
rt(b*sec(f*x + e) + a), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (c + d \sec {\left (e + f x \right )}\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e))**(5/2)*(c+d*sec(f*x+e)),x)
[Out]
Integral((a + b*sec(e + f*x))**(5/2)*(c + d*sec(e + f*x)), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(f*x+e))^(5/2)*(c+d*sec(f*x+e)),x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e) + a)^(5/2)*(d*sec(f*x + e) + c), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(e + f*x))^(5/2)*(c + d/cos(e + f*x)),x)
[Out]
int((a + b/cos(e + f*x))^(5/2)*(c + d/cos(e + f*x)), x)
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