3.10.70 \(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [970]

Optimal. Leaf size=549 \[ -\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^5 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)-16 a^4 C+b^4 (3 A-3 B+C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \]

[Out]

-2/3*(8*a^4*b*B-15*a^2*b^3*B+3*b^5*B-2*a^3*b^2*(A-14*C)+2*a*b^4*(3*A-4*C)-16*a^5*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*(a^3*b*(8*B-12*C)-2*a^2*b^2*(A-3*B-8*C)-3*a*b^3*(A+3*B-3*C)-16*a^4*C+b^4*(
3*A-3*B+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-a*(B*b-C*a))*sec(d*x+c)^2*tan
(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3*a*(4*A*b^4+a*(3*B*a^2*b-7*B*b^3-6*C*a^3+10*C*a*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-B*a*b+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 = 1.29, antiderivative size = 549, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4183, 4175, 4167, 4090, 3917, 4089} \begin {gather*} -\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \cot (c+d x) \left (-16 a^4 C+a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \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 )}-\frac {2 \cot (c+d x) \left (-16 a^5 C+8 a^4 b B-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(-2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 2*a^3*b^2*(A - 14*C) + 2*a*b^4*(3*A - 4*C) - 16*a^5*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))])/(3*b^5*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(a^3*b*(8*B - 12*C) - 2*a^2*b^
2*(A - 3*B - 8*C) - 3*a*b^3*(A + 3*B - 3*C) - 16*a^4*C + b^4*(3*A - 3*B + C))*Cot[c + d*x]*EllipticF[ArcSin[Sq
rt[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))])/(3*b^4*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(A*b^2 - a*(b*B - a*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)) - (2*a*(4*A*b^4 + a*(3*a^2*b*B - 7*b^3*B - 6*a^3*C + 10*a*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 - a*b*B + 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 4183

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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*b*B + 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*(b*B - a*C)*(n - 1)
+ b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(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, B, 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+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac {2 \left (A b^2-a (b B-a 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 (b B-a C)\right )+\frac {3}{2} b (b B-a (A+C)) \sec (c+d x)-\frac {3}{2} \left (A b^2-a b B+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 (b B-a 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 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\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}{4} b \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right )-\frac {1}{4} \left (6 a^4 b B-13 a^2 b^3 B+3 b^5 B+2 a b^4 (2 A-3 C)-12 a^5 C+22 a^3 b^2 C\right ) \sec (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (A b^2-a b B+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 (b B-a 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 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\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-a b B+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 (2 a^3 b B-6 a b^3 B-4 a^4 C+b^4 (3 A+C)+a^2 b^2 (A+7 C)\right )-\frac {3}{8} b \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 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 (b B-a 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 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\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-a b B+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 (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\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 (8 \left (\frac {3}{8} b \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right )-\frac {3}{8} b^2 \left (2 a^3 b B-6 a b^3 B-4 a^4 C+b^4 (3 A+C)+a^2 b^2 (A+7 C)\right )\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 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 (a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)-16 a^4 C+b^4 (3 A-3 B+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 (b B-a 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 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\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-a b B+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 [A]
time = 21.92, size = 989, normalized size = 1.80 \begin {gather*} \frac {4 (b+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left ((a+b) \left (-8 a^4 b B+15 a^2 b^3 B-3 b^5 B+2 a^3 b^2 (A-14 C)+16 a^5 C+2 a b^4 (-3 A+4 C)\right ) E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) \left (-2 a^2 b^2 (A+3 B-8 C)-16 a^4 C+b^4 (3 A+3 B+C)+4 a^3 b (2 B+3 C)+3 a b^3 (A-3 (B+C))\right ) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (-8 a^4 b B+15 a^2 b^3 B-3 b^5 B+2 a^3 b^2 (A-14 C)+16 a^5 C+2 a b^4 (-3 A+4 C)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (b-b \tan ^4\left (\frac {1}{2} (c+d x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2\right )\right )}{3 b^4 \left (a^2-b^2\right )^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {4 \left (2 a^3 A b^2-6 a A b^4-8 a^4 b B+15 a^2 b^3 B-3 b^5 B+16 a^5 C-28 a^3 b^2 C+8 a b^4 C\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right )^2}-\frac {4 \left (a A b^2 \sin (c+d x)-a^2 b B \sin (c+d x)+a^3 C \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}-\frac {4 \left (-a^3 A b^2 \sin (c+d x)+5 a A b^4 \sin (c+d x)+4 a^4 b B \sin (c+d x)-8 a^2 b^3 B \sin (c+d x)-7 a^5 C \sin (c+d x)+11 a^3 b^2 C \sin (c+d x)\right )}{3 b^3 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}+\frac {4 C \tan (c+d x)}{3 b^3}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(4*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*
x)/2]^2)^(-1)]*((a + b)*(-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 2*a^3*b^2*(A - 14*C) + 16*a^5*C + 2*a*b^4*(-3*A
+ 4*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/
2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(a + b)*(-2*a^2*b^2*(A + 3*B - 8
*C) - 16*a^4*C + b^4*(3*A + 3*B + C) + 4*a^3*b*(2*B + 3*C) + 3*a*b^3*(A - 3*(B + C)))*EllipticF[ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*
x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (-8*a^4*b*B + 15*a^2*b^3*B - 3*b^5*B + 2*a^3*b^2*(A - 14*C) + 16*a^
5*C + 2*a*b^4*(-3*A + 4*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(3*b
^4*(a^2 - b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)*(1 + Tan[(c +
d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + ((b +
a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(2*a^3*A*b^2 - 6*a*A*b^4 - 8*a^4*
b*B + 15*a^2*b^3*B - 3*b^5*B + 16*a^5*C - 28*a^3*b^2*C + 8*a*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^2*b*B*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] + 4*a^4*b*B*Sin[c + d*x] - 8*a^2*b^3*B*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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(10851\) vs. \(2(515)=1030\).
time = 0.89, size = 10852, normalized size = 19.77

method	result	size
default	\(\text {Expression too large to display}\)	\(10852\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),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(sec(d*x+c)^3*(A+B*sec(d*x+c)+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 + B*sec(d*x + c)^4 + 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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )} + 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+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/2), 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(sec(d*x+c)^3*(A+B*sec(d*x+c)+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 + B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\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 + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)), x)

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