3.752 \(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=559 \[ \frac {5 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}-\frac {b \left (5 A b^2-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {b \left (-\left (a^4 (3 A-8 C)\right )+26 a^2 A b^2-15 A b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {\left (-\left (a^4 (3 A-8 C)\right )+26 a^2 A b^2-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {\left (6 a^4 C+a^3 b (3 A-2 C)+21 a^2 A b^2-5 a A b^3-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

[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-a^4*(3*A-8*C))*cot(d*x+c)*EllipticE((a+b*se
c(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/(a-b)/b/(a+b)^(3/2)/d+1/3*(21*a^2*A*b^2-5*a*A*b^3-15*A*b^4+a^3*b*(3*A-2*C)+6*a^4*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/(a-b)/b/(a+b)^(3/2)/d+5*A*b*cot(d*x+c)*EllipticPi((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-a^2*(3*A-2*C))*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-1/3*b*(26*a^2*A*b^2-15*A*b^4-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 = 1.19, antiderivative size = 559, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4105, 4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac {b \left (26 a^2 A b^2+a^4 (-(3 A-8 C))-15 A b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {b \left (5 A b^2-a^2 (3 A-2 C)\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (21 a^2 A b^2+a^3 b (3 A-2 C)+6 a^4 C-5 a A b^3-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}-\frac {\left (26 a^2 A b^2+a^4 (-(3 A-8 C))-15 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {5 A b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}+\frac {A \sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((26*a^2*A*b^2 - 15*A*b^4 - 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))])/(3*a^3*(a
- b)*b*(a + b)^(3/2)*d) + ((21*a^2*A*b^2 - 5*a*A*b^3 - 15*A*b^4 + a^3*b*(3*A - 2*C) + 6*a^4*C)*Cot[c + d*x]*El
lipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sq
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*a^3*(a - b)*b*(a + b)^(3/2)*d) + (5*A*b*Sqrt[a + b]*Cot[c + d*x]*Ell
ipticPi[(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 - 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 - 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 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*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)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(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)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3921

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 4004

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 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4058

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 4060

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 4105

Int[((A_.) + 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*(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, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+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 {5 A b}{2}-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-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 {15}{4} A b \left (a^2-b^2\right )+\frac {3}{2} a \left (A b^2+a^2 C\right ) \sec (c+d x)-\frac {1}{4} b \left (5 A b^2-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-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-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 {15}{8} A b \left (a^2-b^2\right )^2+\frac {1}{4} a \left (5 A b^4-3 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-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-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-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 {15}{8} A b \left (a^2-b^2\right )^2+\left (\frac {1}{8} b \left (26 a^2 A b^2-15 A b^4-a^4 (3 A-8 C)\right )+\frac {1}{4} a \left (5 A b^4-3 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-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-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-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-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) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^3}+\frac {\left (21 a^2 A b^2-5 a A b^3-15 A b^4+a^3 b (3 A-2 C)+6 a^4 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-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 (21 a^2 A b^2-5 a A b^3-15 A b^4+a^3 b (3 A-2 C)+6 a^4 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 {5 A b \sqrt {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-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-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]  time = 22.86, size = 1702, normalized size = 3.04 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]*(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*(-5*a^2*A*b^2 + 3*A*b^4 - 2*a^4*C)*Sin[c + d*
x])/(3*a^3*(-a^2 + b^2)^2) + (4*(A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)*(b + a*Cos[c
+ d*x])^2) + (4*(-11*a^2*A*b^3*Sin[c + d*x] + 7*A*b^5*Sin[c + d*x] - 5*a^4*b*C*Sin[c + d*x] + a^2*b^3*C*Sin[c
+ d*x]))/(3*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^
(5/2)) - (2*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)
^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(3*a^5*A*Tan[(c +
d*x)/2] + 3*a^4*A*b*Tan[(c + d*x)/2] - 26*a^3*A*b^2*Tan[(c + d*x)/2] - 26*a^2*A*b^3*Tan[(c + d*x)/2] + 15*a*A*
b^4*Tan[(c + d*x)/2] + 15*A*b^5*Tan[(c + d*x)/2] - 8*a^5*C*Tan[(c + d*x)/2] - 8*a^4*b*C*Tan[(c + d*x)/2] - 6*a
^5*A*Tan[(c + d*x)/2]^3 + 52*a^3*A*b^2*Tan[(c + d*x)/2]^3 - 30*a*A*b^4*Tan[(c + d*x)/2]^3 + 16*a^5*C*Tan[(c +
d*x)/2]^3 + 3*a^5*A*Tan[(c + d*x)/2]^5 - 3*a^4*A*b*Tan[(c + d*x)/2]^5 - 26*a^3*A*b^2*Tan[(c + d*x)/2]^5 + 26*a
^2*A*b^3*Tan[(c + d*x)/2]^5 + 15*a*A*b^4*Tan[(c + d*x)/2]^5 - 15*A*b^5*Tan[(c + d*x)/2]^5 - 8*a^5*C*Tan[(c + d
*x)/2]^5 + 8*a^4*b*C*Tan[(c + d*x)/2]^5 - 30*a^4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]
*Sqrt[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)] + 60*a^2*A*b
^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[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)] - 30*A*b^5*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a
+ b)]*Sqrt[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)] - 30*a
^4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[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)] + 60*a^2*A*b^3*EllipticPi[-1, ArcSin[Ta
n[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[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)] - 30*A*b^5*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*T
an[(c + d*x)/2]^2*Sqrt[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)] + (a + b)*(-26*a^2*A*b^2 + 15*A*b^4 + a^4*(3*A - 8*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)] + 2*a*(a + b)*(3*a*A*b^2 - 5*A*b^3 + 3*a^3*C + a^2*b*(6*A + 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)]))/(3*a*(a^3 - a*b^2)^2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])
^(5/2)*Sqrt[1 + Tan[(c + d*x)/2]^2]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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fricas [F]  time = 43.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)*sec(d*x + c)^2 + A*cos(d*x + c))*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)*cos(d*x + c)/(b*sec(d*x + c) + a)^(5/2), x)

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maple [B]  time = 2.01, size = 6418, normalized size = 11.48 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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