\(\int \frac {\cos ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [577]

Optimal result

Integrand size = 23, antiderivative size = 562 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {\left (33 a^4-170 a^2 b^2+105 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 a^4 (a-b) (a+b)^{3/2} d}+\frac {(a+3 b) \left (6 a^3-45 a^2 b+35 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {\sqrt {a+b} \left (4 a^2+35 b^2\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{4 a^5 d}-\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-7/4*b*sin(d*x+c)/a^2/d/(a+b*sec(d*x+c))^(3/2)+1/2*cos(d*x+c)*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(3/2)-1/12*(33*a
^4-170*a^2*b^2+105*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^4/(a-b)/(a+b)^(3/2)/d+1/12*(a+3*b)*(6*a^3-45*a^2*b+35*
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^4/(a-b)/(a+b)^(3/2)/d-1/4*(4*a^2+35*b^2)*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^5/d-1/12*b^2*(27*a^2-35*b^2)*tan(d*x+c)/a^3/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-1/12*b^2*(
33*a^4-170*a^2*b^2+105*b^4)*tan(d*x+c)/a^4/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 562, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3931, 4189, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {\cos ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}-\frac {\sqrt {a+b} \left (4 a^2+35 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{4 a^5 d}-\frac {\left (33 a^4-170 a^2 b^2+105 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{12 a^4 d (a-b) (a+b)^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {(a+3 b) \left (6 a^3-45 a^2 b+35 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{12 a^4 d (a-b) (a+b)^{3/2}}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}} \]

[In]

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

[Out]

-1/12*((33*a^4 - 170*a^2*b^2 + 105*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))])/(a^4*(a - b)*(a
+ b)^(3/2)*d) + ((a + 3*b)*(6*a^3 - 45*a^2*b + 35*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))])/(
12*a^4*(a - b)*(a + b)^(3/2)*d) - (Sqrt[a + b]*(4*a^2 + 35*b^2)*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))])/(4*a^5*d) - (7*b*Sin[c + d*x])/(4*a^2*d*(a + b*Sec[c + d*x])^(3/2)) + (Cos[c + d*x]*Sin[c
+ d*x])/(2*a*d*(a + b*Sec[c + d*x])^(3/2)) - (b^2*(27*a^2 - 35*b^2)*Tan[c + d*x])/(12*a^3*(a^2 - b^2)*d*(a + b
*Sec[c + d*x])^(3/2)) - (b^2*(33*a^4 - 170*a^2*b^2 + 105*b^4)*Tan[c + d*x])/(12*a^4*(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 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*} \text {integral}& = \frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos (c+d x) \left (-\frac {7 b}{2}+a \sec (c+d x)+\frac {5}{2} b \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a} \\ & = -\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\frac {1}{4} \left (-4 a^2-35 b^2\right )-\frac {5}{2} a b \sec (c+d x)+\frac {21}{4} b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{2 a^2} \\ & = -\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {\int \frac {\frac {3}{8} \left (a^2-b^2\right ) \left (4 a^2+35 b^2\right )+\frac {3}{4} a b \left (3 a^2-7 b^2\right ) \sec (c+d x)-\frac {1}{8} b^2 \left (27 a^2-35 b^2\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a^3 \left (a^2-b^2\right )} \\ & = -\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {-\frac {3}{16} \left (a^2-b^2\right )^2 \left (4 a^2+35 b^2\right )-\frac {1}{8} a b \left (3 a^4-54 a^2 b^2+35 b^4\right ) \sec (c+d x)-\frac {1}{16} b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2} \\ & = -\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {-\frac {3}{16} \left (a^2-b^2\right )^2 \left (4 a^2+35 b^2\right )+\left (-\frac {1}{8} a b \left (3 a^4-54 a^2 b^2+35 b^4\right )+\frac {1}{16} b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (33 a^4-170 a^2 b^2+105 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 a^2+35 b^2\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{8 a^4}+\frac {\left (b (a+3 b) \left (6 a^3-45 a^2 b+35 b^3\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 a^4 (a-b) (a+b)^2} \\ & = -\frac {\left (33 a^4-170 a^2 b^2+105 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}+\frac {(a+3 b) \left (6 a^3-45 a^2 b+35 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}}}{12 a^4 (a-b) (a+b)^{3/2} d}-\frac {\sqrt {a+b} \left (4 a^2+35 b^2\right ) \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}}}{4 a^5 d}-\frac {7 b \sin (c+d x)}{4 a^2 d (a+b \sec (c+d x))^{3/2}}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (27 a^2-35 b^2\right ) \tan (c+d x)}{12 a^3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {b^2 \left (33 a^4-170 a^2 b^2+105 b^4\right ) \tan (c+d x)}{12 a^4 \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. \(1730\) vs. \(2(562)=1124\).

Time = 13.84 (sec) , antiderivative size = 1730, normalized size of antiderivative = 3.08 \[ \int \frac {\cos ^2(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 {2 b^3 \left (-13 a^2+9 b^2\right ) \sin (c+d x)}{3 a^4 \left (-a^2+b^2\right )^2}-\frac {2 b^5 \sin (c+d x)}{3 a^4 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {4 \left (-7 a^2 b^4 \sin (c+d x)+5 b^6 \sin (c+d x)\right )}{3 a^4 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}+\frac {\sin (2 (c+d x))}{4 a^3}\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 {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 (-33 a^5 b \tan \left (\frac {1}{2} (c+d x)\right )-33 a^4 b^2 \tan \left (\frac {1}{2} (c+d x)\right )+170 a^3 b^3 \tan \left (\frac {1}{2} (c+d x)\right )+170 a^2 b^4 \tan \left (\frac {1}{2} (c+d x)\right )-105 a b^5 \tan \left (\frac {1}{2} (c+d x)\right )-105 b^6 \tan \left (\frac {1}{2} (c+d x)\right )+66 a^5 b \tan ^3\left (\frac {1}{2} (c+d x)\right )-340 a^3 b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )+210 a b^5 \tan ^3\left (\frac {1}{2} (c+d x)\right )-33 a^5 b \tan ^5\left (\frac {1}{2} (c+d x)\right )+33 a^4 b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+170 a^3 b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-170 a^2 b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )-105 a b^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )+105 b^6 \tan ^5\left (\frac {1}{2} (c+d x)\right )+24 a^6 \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}}+162 a^4 b^2 \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}}-396 a^2 b^4 \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}}+210 b^6 \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}}+24 a^6 \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}}+162 a^4 b^2 \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}}-396 a^2 b^4 \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}}+210 b^6 \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}}-b \left (33 a^5+33 a^4 b-170 a^3 b^2-170 a^2 b^3+105 a b^4+105 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 \left (6 a^5-3 a^4 b+24 a^3 b^2+54 a^2 b^3-14 a b^4-35 b^5\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 )}{12 a^4 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{5/2} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}{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]^2/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*b^3*(-13*a^2 + 9*b^2)*Sin[c + d*x])/(3*a^4*(-a^2 + b^2)^2) - (2*b^5
*Sin[c + d*x])/(3*a^4*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) - (4*(-7*a^2*b^4*Sin[c + d*x] + 5*b^6*Sin[c + d*x]))
/(3*a^4*(a^2 - b^2)^2*(b + a*Cos[c + d*x])) + Sin[2*(c + d*x)]/(4*a^3)))/(d*(a + b*Sec[c + d*x])^(5/2)) + ((b
+ a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan
[(c + d*x)/2]^2)]*(-33*a^5*b*Tan[(c + d*x)/2] - 33*a^4*b^2*Tan[(c + d*x)/2] + 170*a^3*b^3*Tan[(c + d*x)/2] + 1
70*a^2*b^4*Tan[(c + d*x)/2] - 105*a*b^5*Tan[(c + d*x)/2] - 105*b^6*Tan[(c + d*x)/2] + 66*a^5*b*Tan[(c + d*x)/2
]^3 - 340*a^3*b^3*Tan[(c + d*x)/2]^3 + 210*a*b^5*Tan[(c + d*x)/2]^3 - 33*a^5*b*Tan[(c + d*x)/2]^5 + 33*a^4*b^2
*Tan[(c + d*x)/2]^5 + 170*a^3*b^3*Tan[(c + d*x)/2]^5 - 170*a^2*b^4*Tan[(c + d*x)/2]^5 - 105*a*b^5*Tan[(c + d*x
)/2]^5 + 105*b^6*Tan[(c + d*x)/2]^5 + 24*a^6*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)] + 162*a^4*b^2*Ellipt
icPi[-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)] - 396*a^2*b^4*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)] + 210*b^6*El
lipticPi[-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)] + 24*a^6*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)] + 162*a^4*b^2*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)] - 396*a^2*b^4*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)] + 210*b^6*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)] - b*(33*a^5 + 33*a^4*b - 170*a^3*b^2 - 170*a^2*b^3 + 105*a*b^4 + 105*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*(6*a^5 - 3*a^4*b + 24*a^3*b^2 + 54*a^2*b^3 - 14*a*b^4 - 35*
b^5)*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)]))/(12*a^4*(a^2 - b^2)^2*d*(a + b*Sec[c +
d*x])^(5/2)*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(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. \(7166\) vs. \(2(513)=1026\).

Time = 9.61 (sec) , antiderivative size = 7167, normalized size of antiderivative = 12.75

method	result	size
default	\(\text {Expression too large to display}\)	\(7167\)

[In]

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

[Out]

result too large to display

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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