\(\int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [576]

Optimal result

Integrand size = 21, antiderivative size = 510 \[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) b (a+b)^{3/2} d}+\frac {\left (3 a^3+21 a^2 b-5 a b^2-15 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d}+\frac {5 b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \tan (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)+1/3*(3*a^4-26*a^2*b^2+15*b^4)*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^3/(a-b)/b/
(a+b)^(3/2)/d+1/3*(3*a^3+21*a^2*b-5*a*b^2-15*b^3)*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)/(a+b)^(3/2)/d+5*b*co
t(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*(3*a^2-5*b^2)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec
(d*x+c))^(3/2)+1/3*b*(3*a^4-26*a^2*b^2+15*b^4)*tan(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3931, 4146, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {5 b \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^4 d}+\frac {b \left (3 a^2-5 b^2\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\left (3 a^3+21 a^2 b-5 a b^2-15 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 a^3 d (a-b) (a+b)^{3/2}}+\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^3 b d (a-b) (a+b)^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \tan (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

((3*a^4 - 26*a^2*b^2 + 15*b^4)*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) + ((3*a^3 + 21*a^2*b - 5*a*b^2 - 15*b^3)*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
*(a - b)*(a + b)^(3/2)*d) + (5*b*Sqrt[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) + Sin[c + d*x]/(a*d*(a + b*Sec[c + d*x])^(3/2)) + (b*(3*a^2 - 5*b^2)*Tan[c + d*x])/(3*a^2*(a^2 - b^2)
*d*(a + b*Sec[c + d*x])^(3/2)) + (b*(3*a^4 - 26*a^2*b^2 + 15*b^4)*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 3931

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[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[b*(m + n + 1) - a*(n + 1)*Csc[e + f*x] - b*(m + n + 2)*Csc[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m + 1/2, 0] && ILtQ[n, 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 4146

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(
A*b^2 + 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*b*(A + C)*(m + 1)*Csc[e + f*x] +
(A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Int
egerQ[2*m] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {\int \frac {-\frac {5 b}{2}+\frac {3}{2} b \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{a} \\ & = \frac {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\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} b \left (a^2-b^2\right )-\frac {3}{2} a b^2 \sec (c+d x)-\frac {1}{4} b \left (3 a^2-5 b^2\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 {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\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 (3 a^4-26 a^2 b^2+15 b^4\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} b \left (a^2-b^2\right )^2+\frac {1}{4} a b^2 \left (9 a^2-5 b^2\right ) \sec (c+d x)-\frac {1}{8} b \left (3 a^4-26 a^2 b^2+15 b^4\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 {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\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 (3 a^4-26 a^2 b^2+15 b^4\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} b \left (a^2-b^2\right )^2+\left (\frac {1}{4} a b^2 \left (9 a^2-5 b^2\right )+\frac {1}{8} b \left (3 a^4-26 a^2 b^2+15 b^4\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 (3 a^4-26 a^2 b^2+15 b^4\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 (3 a^4-26 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\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 {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\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 (3 a^4-26 a^2 b^2+15 b^4\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 b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{2 a^3}+\frac {\left (b \left (3 a^3+21 a^2 b-5 a b^2-15 b^3\right )\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 (3 a^4-26 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\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 (3 a^3+21 a^2 b-5 a b^2-15 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\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) (a+b)^{3/2} d}+\frac {5 b \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\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 {\sin (c+d x)}{a d (a+b \sec (c+d x))^{3/2}}+\frac {b \left (3 a^2-5 b^2\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 (3 a^4-26 a^2 b^2+15 b^4\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*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1481\) vs. \(2(510)=1020\).

Time = 14.82 (sec) , antiderivative size = 1481, normalized size of antiderivative = 2.90 \[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {4 b^2 \left (-5 a^2+3 b^2\right ) \sin (c+d x)}{3 a^3 \left (-a^2+b^2\right )^2}+\frac {2 b^4 \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (-11 a^2 b^3 \sin (c+d x)+7 b^5 \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}-\frac {(b+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\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 )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (3 a^5 \tan \left (\frac {1}{2} (c+d x)\right )+3 a^4 b \tan \left (\frac {1}{2} (c+d x)\right )-26 a^3 b^2 \tan \left (\frac {1}{2} (c+d x)\right )-26 a^2 b^3 \tan \left (\frac {1}{2} (c+d x)\right )+15 a b^4 \tan \left (\frac {1}{2} (c+d x)\right )+15 b^5 \tan \left (\frac {1}{2} (c+d x)\right )-6 a^5 \tan ^3\left (\frac {1}{2} (c+d x)\right )+52 a^3 b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )-30 a b^4 \tan ^3\left (\frac {1}{2} (c+d x)\right )+3 a^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a^4 b \tan ^5\left (\frac {1}{2} (c+d x)\right )-26 a^3 b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+26 a^2 b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+15 a b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )-15 b^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )-30 a^4 b \operatorname {EllipticPi}\left (-1,\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 )} \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}}+60 a^2 b^3 \operatorname {EllipticPi}\left (-1,\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 )} \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}}-30 b^5 \operatorname {EllipticPi}\left (-1,\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 )} \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}}-30 a^4 b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\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}}+60 a^2 b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\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}}-30 b^5 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\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 (3 a^5+3 a^4 b-26 a^3 b^2-26 a^2 b^3+15 a b^4+15 b^5\right ) E\left (\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}}+2 a b \left (6 a^3+9 a^2 b-2 a b^2-5 b^3\right ) \operatorname {EllipticF}\left (\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}}\right )}{3 a \left (a^3-a b^2\right )^2 d (a+b \sec (c+d x))^{5/2} \sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-4*b^2*(-5*a^2 + 3*b^2)*Sin[c + d*x])/(3*a^3*(-a^2 + b^2)^2) + (2*b^4
*Sin[c + d*x])/(3*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (2*(-11*a^2*b^3*Sin[c + d*x] + 7*b^5*Sin[c + d*x])
)/(3*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) - ((b + a*Cos[c + d*x])^(5/2)*Se
c[c + d*x]^(5/2)*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*Tan[(c + d*x)/2] + 3*a^4*b*Tan[(c + d*x)/2] - 26*a^3*b^2*Tan[(c + d*x)/2] -
26*a^2*b^3*Tan[(c + d*x)/2] + 15*a*b^4*Tan[(c + d*x)/2] + 15*b^5*Tan[(c + d*x)/2] - 6*a^5*Tan[(c + d*x)/2]^3
+ 52*a^3*b^2*Tan[(c + d*x)/2]^3 - 30*a*b^4*Tan[(c + d*x)/2]^3 + 3*a^5*Tan[(c + d*x)/2]^5 - 3*a^4*b*Tan[(c + d*
x)/2]^5 - 26*a^3*b^2*Tan[(c + d*x)/2]^5 + 26*a^2*b^3*Tan[(c + d*x)/2]^5 + 15*a*b^4*Tan[(c + d*x)/2]^5 - 15*b^5
*Tan[(c + d*x)/2]^5 - 30*a^4*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*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*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*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*b^3*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)] - 30*b^5*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)] + (3*a^5 + 3*a^4*b - 26*a^3
*b^2 - 26*a^2*b^3 + 15*a*b^4 + 15*b^5)*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*
b*(6*a^3 + 9*a^2*b - 2*a*b^2 - 5*b^3)*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 + 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)))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5834\) vs. \(2(469)=938\).

Time = 9.70 (sec) , antiderivative size = 5835, normalized size of antiderivative = 11.44

method	result	size
default	\(\text {Expression too large to display}\)	\(5835\)

[In]

int(cos(d*x+c)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c) + a)*cos(d*x + c)/(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]

\[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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