3.8.48 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [748]
Optimal. Leaf size=488 \[ \frac {4 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^5 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 \left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]
[Out]
4/3*a*(a^2*b^2*(A-14*C)-b^4*(3*A-4*C)+8*a^4*C)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/
(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^5/(a^2-b^2)/d/(a+b)^(1/2)+2/3*(
2*a^2*b^2*(A-8*C)+3*a*b^3*(A-3*C)+16*a^4*C+12*a^3*b*C-b^4*(3*A+C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)
/(a+b)^(1/2),((a+b)/(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^4/(a^2-b^2)
/d/(a+b)^(1/2)-2/3*(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-4/3*a*(2*A*b^4-3
*C*a^4+5*C*a^2*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/3*(A*b^2+2*C*a^2-C*b^2)*(a+b*sec(d*x
+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d
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Rubi [A]
time = 0.90, antiderivative size = 488, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4184, 4175,
4167, 4090, 3917, 4089} \begin {gather*} -\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 C+A b^2-b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {4 a \left (8 a^4 C+a^2 b^2 (A-14 C)-b^4 (3 A-4 C)\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 )}{3 b^5 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \tan (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (16 a^4 C+12 a^3 b C+2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)-b^4 (3 A+C)\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 )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(4*a*(a^2*b^2*(A - 14*C) - b^4*(3*A - 4*C) + 8*a^4*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/S
qrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3
*b^5*Sqrt[a + b]*(a^2 - b^2)*d) + (2*(2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) + 16*a^4*C + 12*a^3*b*C - b^4*(3
*A + C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se
c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(A*b^2 +
a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (4*a*(2*A*b^4 - 3*a^4*C
+ 5*a^2*b^2*C)*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b^2 + 2*a^2*C - b^2*C)*S
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(a^2 - b^2)*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 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]
Rule 4167
Int[csc[(e_.) + (f_.)*(x_)]*((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 + 1)/(b*f*(m
+ 2))), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*
B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 4175
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e
+ f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Rule 4184
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e +
f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1))), x] + Dist[d/(b*(a^2 - b^2)*(m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) + a^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(m
+ n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 -
b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-\frac {3}{2} a b (A+C) \sec (c+d x)-\frac {3}{2} \left (A b^2+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{2} b \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right )-\frac {1}{2} a \left (2 A b^4-\left (6 a^4-11 a^2 b^2+3 b^4\right ) C\right ) \sec (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (A b^2+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (\frac {3}{8} b^2 \left (4 a^4 C-b^4 (3 A+C)-a^2 b^2 (A+7 C)\right )+\frac {3}{4} a b \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}+\frac {\left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b^3 (a+b)^2}\\ &=\frac {4 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\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}}}{3 (a-b) b^5 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\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}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4050\) vs. \(2(488)=976\).
time = 26.56, size = 4050, normalized size = 8.30 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-8*a*(a^2*A*b^2 - 3*A*b^4 + 8*a^4*C - 14*a^2*b^2
*C + 4*b^4*C)*Sin[c + d*x])/(3*b^4*(a^2 - b^2)^2) - (4*(a*A*b^2*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(3*b^2*(-a
^2 + b^2)*(b + a*Cos[c + d*x])^2) - (4*(-(a^3*A*b^2*Sin[c + d*x]) + 5*a*A*b^4*Sin[c + d*x] - 7*a^5*C*Sin[c + d
*x] + 11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (4*C*Tan[c + d*x])/(3*b^3)))/(
d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (8*(b + a*Cos[c + d*x])^2*((4*a^3*A)/(3*b*(-a^2
+ b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (4*a*A*b)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sq
rt[Sec[c + d*x]]) + (32*a^5*C)/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (56*a^3*C)
/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a*b*C)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Co
s[c + d*x]]*Sqrt[Sec[c + d*x]]) - (10*a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) +
(4*a^4*A*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/((
-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (10*a^2*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x
]]) + (32*a^6*C*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (64*a^4*C*Sqrt[Sec[c + d
*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*b^2*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b
+ a*Cos[c + d*x]]) - (4*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) +
(4*a^4*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (16*a^2*C*Cos
[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (32*a^6*C*Cos[2*(c + d*x)]*Sqr
t[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (56*a^4*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x
]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*
(A + C*Sec[c + d*x]^2)*(2*a*(a + b)*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*Sqrt[Cos[c + d*x]/(1 + C
os[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a -
b)/(a + b)] + b*(a + b)*(-2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) - 16*a^4*C + 12*a^3*b*C + b^4*(3*A + C))*Sq
rt[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[T
an[(c + d*x)/2]], (a - b)/(a + b)] + a*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*Cos[c + d*x]*(b + a*C
os[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(a^2 - b^2)^2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[
Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2)*((4*a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*a*(
a + b)*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(a + b)*(-2*
a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) - 16*a^4*C + 12*a^3*b*C + b^4*(3*A + C))*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)] + a*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Tan[(c + d*x)/2]))/(3*b^4*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) - (4*Sqrt[Cos[
(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*a*(a + b)*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*S
qrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[
Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(a + b)*(-2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) - 16*a^4*C + 12*a^3*
b*C + b^4*(3*A + C))*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)] + a*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C)
)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^4*(a^2 - b^2)^2*Sqrt[b + a*Cos[
c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (8*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*((a*(a^2*b^2*(A - 14*C) + 8*a^4
*C + b^4*(-3*A + 4*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4)/2 + (a*(a + b)*(a^2*b^2*(A - 14*C
) + 8*a^4*C + b^4*(-3*A + 4*C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(
c + d*x)/2]], (a - b)/(a + b)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d
*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + (b*(a + b)*(-2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) - 16*a^4*C
+ 12*a^3*b*C + b^4*(3*A + C))*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)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*
x])))/(2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]) + (a*(a + b)*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))
*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Elliptic...
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(7050\) vs.
\(2(454)=908\).
time = 0.61, size = 7051, normalized size = 14.45
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method |
result |
size |
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default |
\(\text {Expression too large to display}\) |
\(7051\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
result too large to display
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Maxima [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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^5 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x
+ c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/2), x)
________________________________________________________________________________________
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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/2), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)),x)
[Out]
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)), x)
________________________________________________________________________________________