3.10.74 \(\int \frac {\cos (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [974]

Optimal. Leaf size=618 \[ -\frac {\left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) b (a+b)^{3/2} d}-\frac {\left (15 A b^4+a b^3 (5 A-6 B)-a^2 b^2 (21 A+2 B)-6 a^4 C-a^3 b (3 A-2 (6 B+C))\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 b \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

A*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)-1/3*(26*a^2*A*b^2-15*A*b^4-14*a^3*b*B+6*a*b^3*B-a^4*(3*A-8*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+se
c(d*x+c))/(a-b))^(1/2)/a^3/(a-b)/b/(a+b)^(3/2)/d-1/3*(15*A*b^4+a*b^3*(5*A-6*B)-a^2*b^2*(21*A+2*B)-6*a^4*C-a^3*
b*(3*A-12*B-2*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)/a^3/b/(a^2-b^2)/d/(a+b)^(1/2)+(5*A*b-2*B*a)*cot(d*x+c)*Ellipti
cPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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)/a^4/d-1/3*b*(5*A*b^2-2*a*b*B-a^2*(3*A-2*C))*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*s
ec(d*x+c))^(3/2)-1/3*b*(26*a^2*A*b^2-15*A*b^4-14*a^3*b*B+6*a*b^3*B-a^4*(3*A-8*C))*tan(d*x+c)/a^3/(a^2-b^2)^2/d
/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 1.02, antiderivative size = 618, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4189, 4145, 4143, 4006, 3869, 3917, 4089} \begin {gather*} \frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}-\frac {b \tan (c+d x) \left (-\left (a^2 (3 A-2 C)\right )-2 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\cot (c+d x) \left (-6 a^4 C-a^3 b (3 A-2 (6 B+C))-a^2 b^2 (21 A+2 B)+a b^3 (5 A-6 B)+15 A b^4\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 a^3 b d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\cot (c+d x) \left (-\left (a^4 (3 A-8 C)\right )-14 a^3 b B+26 a^2 A b^2+6 a b^3 B-15 A b^4\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 a^3 b d (a-b) (a+b)^{3/2}}-\frac {b \tan (c+d x) \left (-\left (a^4 (3 A-8 C)\right )-14 a^3 b B+26 a^2 A b^2+6 a b^3 B-15 A b^4\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((26*a^2*A*b^2 - 15*A*b^4 - 14*a^3*b*B + 6*a*b^3*B - a^4*(3*A - 8*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))])/(a^3*(a - b)*b*(a + b)^(3/2)*d) - ((15*A*b^4 + a*b^3*(5*A - 6*B) - a^2*b^2*(21*A + 2*B) - 6
*a^4*C - a^3*b*(3*A - 2*(6*B + 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))])/(3*a^3*b*Sqrt[a + b]
*(a^2 - b^2)*d) + (Sqrt[a + b]*(5*A*b - 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))
])/(a^4*d) + (A*Sin[c + d*x])/(a*d*(a + b*Sec[c + d*x])^(3/2)) - (b*(5*A*b^2 - 2*a*b*B - a^2*(3*A - 2*C))*Tan[
c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (b*(26*a^2*A*b^2 - 15*A*b^4 - 14*a^3*b*B + 6*a*b^
3*B - a^4*(3*A - 8*C))*Tan[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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 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 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]

Rule 4189

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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cos (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 {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\frac {1}{2} (5 A b-2 a B)-a C \sec (c+d x)-\frac {3}{2} A b \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{a}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {-\frac {3}{4} \left (a^2-b^2\right ) (5 A b-2 a B)+\frac {3}{2} a \left (A b^2-a (b B-a C)\right ) \sec (c+d x)-\frac {1}{4} b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\frac {3}{8} \left (a^2-b^2\right )^2 (5 A b-2 a B)+\frac {1}{4} a \left (5 A b^4+6 a^3 b B-2 a b^3 B-3 a^4 C-a^2 b^2 (9 A+C)\right ) \sec (c+d x)-\frac {1}{8} b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\frac {3}{8} \left (a^2-b^2\right )^2 (5 A b-2 a B)+\left (\frac {1}{8} b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right )+\frac {1}{4} a \left (5 A b^4+6 a^3 b B-2 a b^3 B-3 a^4 C-a^2 b^2 (9 A+C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}+\frac {\left (b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 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^3 (a-b) b (a+b)^{3/2} d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {(5 A b-2 a B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^3}-\frac {\left (15 A b^4+a b^3 (5 A-6 B)-a^2 b^2 (21 A+2 B)-6 a^4 C-a^3 b (3 A-2 (6 B+C))\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{6 a^3 (a-b) (a+b)^2}\\ &=-\frac {\left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 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^3 (a-b) b (a+b)^{3/2} d}-\frac {\left (15 A b^4+a b^3 (5 A-6 B)-a^2 b^2 (21 A+2 B)-6 a^4 C-a^3 b (3 A-2 (6 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^3 (a-b) b (a+b)^{3/2} d}+\frac {\sqrt {a+b} (5 A b-2 a B) \cot (c+d x) \Pi \left (\frac {a+b}{a};\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}}}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (5 A b^2-2 a b B-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b \left (26 a^2 A b^2-15 A b^4-14 a^3 b B+6 a b^3 B-a^4 (3 A-8 C)\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 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. \(20027\) vs. \(2(618)=1236\).
time = 29.20, size = 20027, normalized size = 32.41 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(10318\) vs. \(2(576)=1152\).
time = 0.36, size = 10319, normalized size = 16.70

method	result	size
default	\(\text {Expression too large to display}\)	\(10319\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(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(cos(d*x+c)*(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(-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(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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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(cos(d*x+c)*(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)*cos(d*x + c)/(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 {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\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((cos(c + d*x)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(5/2),x)

[Out]

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

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