\(\int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\) [758]

Optimal result

Integrand size = 32, antiderivative size = 445 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {2 \left (11 a^2-3 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a (a-b) (a+b)^{3/2} d}-\frac {2 \left (9 a^2-2 a b-3 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a (a-b) (a+b)^{3/2} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

2/3*(11*a^2-3*b^2)*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/(a-b)/(a+b)^(3/2)/d-2/3*(9*a^2-2*a*b-3*b^2)*cot(d*x+c)*Elli
pticF((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/(a-b)/(a+b)^(3/2)/d-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^2/d+4/3*b^2*tan(d*
x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*b^2*(11*a^2-3*b^2)*tan(d*x+c)/a/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/
2)

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4127, 4008, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\frac {2 \left (9 a^2-2 a b-3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 a d (a-b) (a+b)^{3/2}}+\frac {2 \left (11 a^2-3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a d (a-b) (a+b)^{3/2}}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {4 b^2 \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}} \]

[In]

Int[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

(2*(11*a^2 - 3*b^2)*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*(a - b)*(a + b)^(3/2)*d) - (2*(
9*a^2 - 2*a*b - 3*b^2)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*a*(a - b)*(a + b)^(3/2)*d) - (
2*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^2*d) + (4*b^2*Tan[c + d*x])
/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*b^2*(11*a^2 - 3*b^2)*Tan[c + d*x])/(3*a*(a^2 - b^2)^2*d*Sqr
t[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 4008

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

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 4127

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

Rubi steps \begin{align*} \text {integral}& = -\int \frac {-a+b \sec (c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx \\ & = \frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {\frac {3}{2} a \left (a^2-b^2\right )-3 a^2 b \sec (c+d x)+a b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {-\frac {3}{4} a \left (a^2-b^2\right )^2+\frac {1}{4} a^2 b \left (9 a^2-b^2\right ) \sec (c+d x)+\frac {1}{4} a b^2 \left (11 a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {-\frac {3}{4} a \left (a^2-b^2\right )^2+\left (-\frac {1}{4} a b^2 \left (11 a^2-3 b^2\right )+\frac {1}{4} a^2 b \left (9 a^2-b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}-\frac {\left (b^2 \left (11 a^2-3 b^2\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (11 a^2-3 b^2\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 (a-b) (a+b)^{3/2} d}+\frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a}-\frac {\left (b \left (9 a^2-2 a b-3 b^2\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a (a-b) (a+b)^2} \\ & = \frac {2 \left (11 a^2-3 b^2\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 (a-b) (a+b)^{3/2} d}-\frac {2 \left (9 a^2-2 a b-3 b^2\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 (a-b) (a+b)^{3/2} d}-\frac {2 \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^2 d}+\frac {4 b^2 \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (11 a^2-3 b^2\right ) \tan (c+d x)}{3 a \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. \(1434\) vs. \(2(445)=890\).

Time = 15.05 (sec) , antiderivative size = 1434, normalized size of antiderivative = 3.22 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^2(c+d x) (a-b \sec (c+d x)) \left (\frac {2 b \left (-11 a^2+3 b^2\right ) \sin (c+d x)}{3 a \left (-a^2+b^2\right )^2}-\frac {4 b^3 \sin (c+d x)}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (-13 a^2 b^2 \sin (c+d x)+5 b^4 \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (-b+a \cos (c+d x)) (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x) (a-b \sec (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 (11 a^3 b \tan \left (\frac {1}{2} (c+d x)\right )+11 a^2 b^2 \tan \left (\frac {1}{2} (c+d x)\right )-3 a b^3 \tan \left (\frac {1}{2} (c+d x)\right )-3 b^4 \tan \left (\frac {1}{2} (c+d x)\right )-22 a^3 b \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 a b^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )+11 a^3 b \tan ^5\left (\frac {1}{2} (c+d x)\right )-11 a^2 b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+6 a^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}}-12 a^2 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}}+6 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}}+6 a^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}}-12 a^2 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}}+6 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}}+b \left (11 a^3+11 a^2 b-3 a b^2-3 b^3\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}}-a \left (3 a^3+9 a^2 b+5 a b^2-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^2-b^2\right )^2 d (-b+a \cos (c+d x)) (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[(a^2 - b^2*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^2*(a - b*Sec[c + d*x])*((2*b*(-11*a^2 + 3*b^2)*Sin[c + d*x])/(3*a*(-a^2 +
b^2)^2) - (4*b^3*Sin[c + d*x])/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) - (2*(-13*a^2*b^2*Sin[c + d*x] + 5*b^
4*Sin[c + d*x]))/(3*a*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(-b + a*Cos[c + d*x])*(a + b*Sec[c + d*x])^(5/2
)) + (2*(b + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)*(a - b*Sec[c + d*x])*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2
+ b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(11*a^3*b*Tan[(c + d*x)/2] + 11*a^2*b^2*Tan[(c + d*x)/2] - 3
*a*b^3*Tan[(c + d*x)/2] - 3*b^4*Tan[(c + d*x)/2] - 22*a^3*b*Tan[(c + d*x)/2]^3 + 6*a*b^3*Tan[(c + d*x)/2]^3 +
11*a^3*b*Tan[(c + d*x)/2]^5 - 11*a^2*b^2*Tan[(c + d*x)/2]^5 - 3*a*b^3*Tan[(c + d*x)/2]^5 + 3*b^4*Tan[(c + d*x)
/2]^5 + 6*a^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)] - 12*a^2*b^2*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)] + 6*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)] + 6*a^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)] - 12*a^2*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)] + 6*b^4*Ell
ipticPi[-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*(11*a^3 + 11*a^2*b - 3*a*b^2 - 3*b^3)*Ellipti
cE[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)] - a*(3*a^3 + 9*a^2*b + 5*a*b^2 - b^3)*EllipticF[ArcS
in[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*T
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(3*a*(a^2 - b^2)^2*d*(-b + a*Cos[c + d*x])*(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. \(5001\) vs. \(2(406)=812\).

Time = 11.45 (sec) , antiderivative size = 5002, normalized size of antiderivative = 11.24

method	result	size
default	\(\text {Expression too large to display}\)	\(5002\)
parts	\(\text {Expression too large to display}\)	\(14038\)

[In]

int((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { -\frac {b^{2} \sec \left (d x + c\right )^{2} - a^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sec(d*x + c) + a)*(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)

Sympy [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {a - b \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a**2-b**2*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(7/2),x)

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { -\frac {b^{2} \sec \left (d x + c\right )^{2} - a^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a^2-b^2*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(-(b^2*sec(d*x + c)^2 - a^2)/(b*sec(d*x + c) + a)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\int -\frac {a^2-\frac {b^2}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

[In]

int((a^2 - b^2/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(7/2),x)

[Out]

-int(-(a^2 - b^2/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(7/2), x)