3.6.45 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\) [545]
Optimal. Leaf size=399 \[ \frac {2 (a-b) \sqrt {a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d} \]
[Out]
2/315*(a-b)*(10*a^4-279*a^2*b^2-147*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))
^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d+2/315*(a-b)*(10*a^3+1
65*a^2*b-114*a*b^2+147*b^3)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)
^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d-2/315*(10*a^2-49*b^2)*(a+b*sec(d*x
+c))^(3/2)*tan(d*x+c)/b/d-4/63*a*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*(a+b*sec(d*x+c))^(7/2)*tan(d*x+c)/b
/d-4/315*a*(5*a^2-57*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
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Rubi [A]
time = 0.52, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3925, 4087,
4090, 3917, 4089} \begin {gather*} -\frac {2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b d}-\frac {4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{315 b d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^2 d}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(2*(a - b)*Sqrt[a + b]*(10*a^4 - 279*a^2*b^2 - 147*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(315*b^3*d) + (2*(a - b)*Sqrt[a + b]*(10*a^3 + 165*a^2*b - 114*a*b^2 + 147*b^3)*Cot[c + d*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[c + d*x]))/(a - b))])/(315*b^2*d) - (4*a*(5*a^2 - 57*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(315*b*d) -
(2*(10*a^2 - 49*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*b*d) - (4*a*(a + b*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(63*b*d) + (2*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d)
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 3925
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*
(b*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Rule 4087
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[Csc[e + f
*x]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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 4090
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, 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}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {2 \int \sec (c+d x) \left (\frac {7 b}{2}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{5/2} \, dx}{9 b}\\ &=-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {39 a b}{4}-\frac {1}{4} \left (10 a^2-49 b^2\right ) \sec (c+d x)\right ) \, dx}{63 b}\\ &=-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {8 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {3}{8} b \left (55 a^2+49 b^2\right )-\frac {3}{4} a \left (5 a^2-57 b^2\right ) \sec (c+d x)\right ) \, dx}{315 b}\\ &=-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {16 \int \frac {\sec (c+d x) \left (\frac {3}{16} a b \left (155 a^2+261 b^2\right )-\frac {3}{16} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b}\\ &=-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {\left ((a-b) \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b}-\frac {\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b}\\ &=\frac {2 (a-b) \sqrt {a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}-\frac {4 a \left (5 a^2-57 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac {2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac {4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}\\ \end {align*}
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Mathematica [A]
time = 16.56, size = 552, normalized size = 1.38 \begin {gather*} \frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 \left (10 a^5+10 a^4 b-279 a^3 b^2-279 a^2 b^3-147 a b^4-147 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b \left (-10 a^4+155 a^3 b+279 a^2 b^2+261 a b^3+147 b^4\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+\left (10 a^4-279 a^2 b^2-147 b^4\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^2 d (b+a \cos (c+d x))^3 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {2 \left (-10 a^4+279 a^2 b^2+147 b^4\right ) \sin (c+d x)}{315 b^2}+\frac {2}{315} \sec ^2(c+d x) \left (75 a^2 \sin (c+d x)+49 b^2 \sin (c+d x)\right )+\frac {2 \sec (c+d x) \left (5 a^3 \sin (c+d x)+163 a b^2 \sin (c+d x)\right )}{315 b}+\frac {38}{63} a b \sec ^2(c+d x) \tan (c+d x)+\frac {2}{9} b^2 \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*(10*a^5 + 10*a^4*b - 279*a^3*b^2 - 279*
a^2*b^3 - 147*a*b^4 - 147*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + C
os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(-10*a^4 + 155*a^3*b + 279*a^2*b^2 +
261*a*b^3 + 147*b^4)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*
x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (10*a^4 - 279*a^2*b^2 - 147*b^4)*Cos[c + d*x]*(b
+ a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c + d*x)/2
]^2]*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(-10*a^4 + 279*a^2*b^2 + 147*b^4)*Si
n[c + d*x])/(315*b^2) + (2*Sec[c + d*x]^2*(75*a^2*Sin[c + d*x] + 49*b^2*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(
5*a^3*Sin[c + d*x] + 163*a*b^2*Sin[c + d*x]))/(315*b) + (38*a*b*Sec[c + d*x]^2*Tan[c + d*x])/63 + (2*b^2*Sec[c
+ d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2522\) vs.
\(2(361)=722\).
time = 0.53, size = 2523, normalized size = 6.32
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\(\text {Expression too large to display}\) |
\(2523\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^3*(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
-2/315/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(5*cos(d*x+c)^6*a^4*b+279*cos(
d*x+c)^6*a^3*b^2+163*cos(d*x+c)^6*a^2*b^3+147*cos(d*x+c)^6*a*b^4-10*cos(d*x+c)^5*a^4*b-199*cos(d*x+c)^5*a^3*b^
2+279*cos(d*x+c)^5*a^2*b^3+65*cos(d*x+c)^5*a*b^4+5*cos(d*x+c)^4*a^4*b-272*cos(d*x+c)^4*a^2*b^3-80*cos(d*x+c)^3
*a^3*b^2-82*cos(d*x+c)^3*a*b^4-170*cos(d*x+c)^2*a^2*b^3-130*cos(d*x+c)*a*b^4+147*cos(d*x+c)^5*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a
+b))^(1/2))*sin(d*x+c)*b^5+10*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^5-147*cos(d*x+c)^5*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),
((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^5+147*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^5+10*cos(d*x+c)^
4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^5-147*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^5+155*c
os(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b^2+279*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*
x+c)*a^2*b^3+261*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*
EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^4+10*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b)
)^(1/2))*sin(d*x+c)*a^4*b-279*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b^2-279*cos(d*x+c)^4*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^3-147*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^4-35*b
^5-10*cos(d*x+c)^6*a^5+10*cos(d*x+c)^5*a^5+147*cos(d*x+c)^5*b^5-98*cos(d*x+c)^4*b^5-14*cos(d*x+c)^2*b^5-10*cos
(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*
x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4*b+155*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)
*a^3*b^2+279*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elli
pticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^3+261*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))
^(1/2))*sin(d*x+c)*a*b^4+10*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4*b-279*cos(d*x+c)^5*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),
((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b^2-279*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(
1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^3-147*co
s(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d
*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^4-10*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)
*a^4*b)/(b+a*cos(d*x+c))/cos(d*x+c)^4/sin(d*x+c)^5/b^2
________________________________________________________________________________________
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(sec(d*x+c)^3*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^3, 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(sec(d*x+c)^3*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((b^2*sec(d*x + c)^5 + 2*a*b*sec(d*x + c)^4 + a^2*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x)
________________________________________________________________________________________
Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
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(sec(d*x+c)^3*(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^3, x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^3,x)
[Out]
int((a + b/cos(c + d*x))^(5/2)/cos(c + d*x)^3, x)
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