3.9.54 \(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\) [854]

Optimal. Leaf size=446 \[ -\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b \sqrt {a+b} \left (a^2-b^2\right )^2 d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/15*(23*B*a^2*b+9*B*b^3-3*C*a^3-29*C*a*b^2)*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)/(a-b)^2/b^2/(a+b)^(5/2)/d+2/15*(3*
a^2*(5*B+C)-8*a*b*(B+3*C)+b^2*(9*B+5*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/(a^2-b^2)^2/d/(a+b)^(1/2)-2/5*(B*b-C*
a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(5/2)-2/15*(8*B*a*b-3*C*a^2-5*C*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*
sec(d*x+c))^(3/2)-2/15*(23*B*a^2*b+9*B*b^3-3*C*a^3-29*C*a*b^2)*tan(d*x+c)/(a^2-b^2)^3/d/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.57, antiderivative size = 446, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4145, 4143, 12, 3917, 4089} \begin {gather*} \frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 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 )}{15 b d \sqrt {a+b} \left (a^2-b^2\right )^2}-\frac {2 \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}-\frac {2 (b B-a C) \tan (c+d x)}{5 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\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 )}{15 b^2 d (a-b)^2 (a+b)^{5/2}}-\frac {2 \left (-3 a^3 C+23 a^2 b B-29 a b^2 C+9 b^3 B\right ) \tan (c+d x)}{15 d \left (a^2-b^2\right )^3 \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

(-2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*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))])/(15*(a
- b)^2*b^2*(a + b)^(5/2)*d) + (2*(3*a^2*(5*B + C) - 8*a*b*(B + 3*C) + b^2*(9*B + 5*C))*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 + Sec[c + d*x]))/(a - b))])/(15*b*Sqrt[a + b]*(a^2 - b^2)^2*d) - (2*(b*B - a*C)*Tan[c + d*x])/(5*(a^2 - b^
2)*d*(a + b*Sec[c + d*x])^(5/2)) - (2*(8*a*b*B - 3*a^2*C - 5*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^2*d*(a + b*S
ec[c + d*x])^(3/2)) - (2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Tan[c + d*x])/(15*(a^2 - b^2)^3*d*Sqrt[
a + b*Sec[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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
]

Rule 4145

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)
*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 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]

Rubi steps

\begin {align*} \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx &=-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} a (a B-b C) \sec (c+d x)+\frac {3}{2} a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{5 a \left (a^2-b^2\right )}\\ &=-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} a^2 \left (5 a^2 B+3 b^2 B-8 a b C\right ) \sec (c+d x)-\frac {1}{4} a^2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{15 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {-\frac {1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right ) \sec (c+d x)-\frac {1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {\left (\frac {1}{8} a^3 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right )-\frac {1}{8} a^3 \left (15 a^3 B+17 a b^2 B-27 a^2 b C-5 b^3 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}+\frac {\left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^3}\\ &=-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 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}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 (a-b)^2 (a+b)^3}\\ &=-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 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}}}{15 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac {2 \left (3 a^2 (5 B+C)-8 a b (B+3 C)+b^2 (9 B+5 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}}}{15 (a-b)^2 b (a+b)^{5/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{5 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (8 a b B-3 a^2 C-5 b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (23 a^2 b B+9 b^3 B-3 a^3 C-29 a b^2 C\right ) \tan (c+d x)}{15 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(3729\) vs. \(2(446)=892\).
time = 24.64, size = 3729, normalized size = 8.36 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

((b + a*Cos[c + d*x])^4*Sec[c + d*x]^4*((-2*(23*a^2*b*B + 9*b^3*B - 3*a^3*C - 29*a*b^2*C)*Sin[c + d*x])/(15*b*
(-a^2 + b^2)^3) - (2*(b^3*B*Sin[c + d*x] - a*b^2*C*Sin[c + d*x]))/(5*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^3) -
(2*(-14*a^2*b^2*B*Sin[c + d*x] + 6*b^4*B*Sin[c + d*x] + 9*a^3*b*C*Sin[c + d*x] - a*b^3*C*Sin[c + d*x]))/(15*a
^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (2*(-34*a^4*b*B*Sin[c + d*x] + 5*a^2*b^3*B*Sin[c + d*x] - 3*b^5*B*S
in[c + d*x] + 9*a^5*C*Sin[c + d*x] + 25*a^3*b^2*C*Sin[c + d*x] - 2*a*b^4*C*Sin[c + d*x]))/(15*a^2*(a^2 - b^2)^
3*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(7/2)) - (2*(b + a*Cos[c + d*x])^3*((23*a^2*b*B)/(15*(-a^2 +
b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (3*b^3*B)/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sq
rt[Sec[c + d*x]]) - (a^3*C)/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (29*a*b^2*C)/(15*
(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a^3*B*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*
Sqrt[b + a*Cos[c + d*x]]) - (8*a*b^2*B*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (a^4
*C*Sqrt[Sec[c + d*x]])/(5*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (2*a^2*b*C*Sqrt[Sec[c + d*x]])/(15*(-a^
2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) + (b^3*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]])
+ (23*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) + (3*a*b^2*B*Cos
[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (a^4*C*Cos[2*(c + d*x)]*Sqrt[S
ec[c + d*x]])/(5*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]) - (29*a^2*b*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])
/(15*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]))*Sec[c + d*x]^(7/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a
+ b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*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)] + 2*b*(a + b)*(b^2*(9
*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - 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)] + (-23*a^2*b*B - 9*b^3*B +
3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b*(-a^2 + b
^2)^3*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(7/2)*(-1/15*(a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Si
n[c + d*x]*(2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqr
t[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*
b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - 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)] + (-23*a^
2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])
)/(b*(-a^2 + b^2)^3*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*
x]]*Tan[(c + d*x)/2]*(2*(a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*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)] + 2*b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - 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)]
+ (-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c
+ d*x)/2]))/(15*b*(-a^2 + b^2)^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - (2*Sqrt[Cos[(c + d*x)/2]
^2*Sec[c + d*x]]*(((-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d
*x)/2]^4)/2 + ((a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*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)*(b^2*(9*B - 5*
C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B - C))*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcS
in[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 + C
os[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] + ((a + b)*(-23*a^2*b*B - 9*b^3*B + 3*a^3*C + 29*a*b^2*C)
*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x]
)/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b
+ a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] + (b*(a + b)*(b^2*(9*B - 5*C) + 8*a*b*(B - 3*C) + 3*a^2*(5*B
- C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c +
d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*...

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(7694\) vs. \(2(412)=824\).
time = 0.29, size = 7695, normalized size = 17.25

method	result	size
default	\(\text {Expression too large to display}\)	\(7695\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x +
c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(7/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**(7/2), 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((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(b*sec(d*x + c) + a)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(7/2),x)

[Out]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(7/2), x)

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