3.7.14 \(\int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [614]

Optimal. Leaf size=342 \[ \frac {a \left (5 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a^2 \left (5 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

1/3*(5*a^2-2*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/b^2/(a^2-b^2)/d-a^2*sec(d*x+c)^(5/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a
+b*sec(d*x+c))-a*(5*a^2-4*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d+a*(5*a^2-4*b^2)*(cos(1/2*d*x+1/2*c)
^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a^2-
b^2)/d+1/3*(5*a^2-2*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))
*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^2-b^2)/d+a^2*(5*a^2-7*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+
1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^3/(a+b)^2/d

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Rubi [A]
time = 0.63, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3930, 4187, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\left (5 a^2-2 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b^2 d \left (a^2-b^2\right )}+\frac {\left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {a \left (5 a^2-4 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{b^3 d \left (a^2-b^2\right )}+\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac {a^2 \left (5 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(9/2)/(a + b*Sec[c + d*x])^2,x]

[Out]

(a*(5*a^2 - 4*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a^2 - b^2)*d) + ((5*
a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^2*(a^2 - b^2)*d) + (a^2*(5*
a^2 - 7*b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^3*(a
+ b)^2*d) - (a*(5*a^2 - 4*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) + ((5*a^2 - 2*b^2)*Sec[c +
d*x]^(3/2)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) - (a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a +
b*Sec[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3930

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)
*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist
[d^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b
*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 3934

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4187

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[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(
m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(
d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a
*C*n)*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] && GtQ[n, 0]

Rule 4191

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a^2}{2}-a b \sec (c+d x)-\frac {1}{2} \left (5 a^2-2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {2 \int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a \left (5 a^2-2 b^2\right )+\frac {1}{2} b \left (2 a^2+b^2\right ) \sec (c+d x)+\frac {3}{4} a \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {-\frac {3}{8} a^2 \left (5 a^2-4 b^2\right )-\frac {1}{4} a b \left (10 a^2-7 b^2\right ) \sec (c+d x)-\frac {1}{8} \left (15 a^4-16 a^2 b^2-2 b^4\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {4 \int \frac {-\frac {3}{8} a^3 \left (5 a^2-4 b^2\right )-\left (\frac {1}{4} a^2 b \left (10 a^2-7 b^2\right )-\frac {3}{8} a^2 b \left (5 a^2-4 b^2\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^2 b^3 \left (a^2-b^2\right )}+\frac {\left (a^2 \left (5 a^2-7 b^2\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (a \left (5 a^2-4 b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (5 a^2-2 b^2\right ) \int \sqrt {\sec (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )}+\frac {\left (a^2 \left (5 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac {a^2 \left (5 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (a \left (5 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (\left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}\\ &=\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d}+\frac {a^2 \left (5 a^2-7 b^2\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}-\frac {a \left (5 a^2-4 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (5 a^2-2 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 35.04, size = 294, normalized size = 0.86 \begin {gather*} \frac {\frac {2 b \left (-\frac {3 a^2 \left (5 a^2-4 b^2\right ) \sin (c+d x)}{a^2-b^2}+2 b (-5 a+b \sec (c+d x)) \tan (c+d x)\right )}{(b+a \cos (c+d x)) \sqrt {\sec (c+d x)}}+\frac {\cot (c+d x) \left (-6 a b \left (5 a^2-4 b^2\right ) E\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 \left (15 a^4+15 a^3 b-16 a^2 b^2-12 a b^3-2 b^4\right ) F\left (\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-6 a \left (b \left (-5 a^2+4 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin ^2(c+d x)+a \left (5 a^2-7 b^2\right ) \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}\right )\right )}{(a-b) (a+b)}}{6 b^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(9/2)/(a + b*Sec[c + d*x])^2,x]

[Out]

((2*b*((-3*a^2*(5*a^2 - 4*b^2)*Sin[c + d*x])/(a^2 - b^2) + 2*b*(-5*a + b*Sec[c + d*x])*Tan[c + d*x]))/((b + a*
Cos[c + d*x])*Sqrt[Sec[c + d*x]]) + (Cot[c + d*x]*(-6*a*b*(5*a^2 - 4*b^2)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]]
, -1]*Sqrt[-Tan[c + d*x]^2] + 2*(15*a^4 + 15*a^3*b - 16*a^2*b^2 - 12*a*b^3 - 2*b^4)*EllipticF[ArcSin[Sqrt[Sec[
c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 6*a*(b*(-5*a^2 + 4*b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x]^2 + a*(5*a^2 -
7*b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2])))/((a - b)*(a + b)))/(6*b^4*
d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(974\) vs. \(2(398)=796\).
time = 0.42, size = 975, normalized size = 2.85

method	result	size
default	\(\text {Expression too large to display}\)	\(975\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(9/2)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^2/b^2*(1/b*a^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))+1/2/b*a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b*a/(a^2-b^2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*
a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2/b^2*(-1
/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-4*a^3/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2
^(1/2))-4/b^3*a/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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(sec(d*x+c)^(9/2)/(a+b*sec(d*x+c))^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(sec(d*x+c)^(9/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(9/2)/(a+b*sec(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7318 deep

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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(sec(d*x+c)^(9/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(9/2)/(b*sec(d*x + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(9/2)/(a + b/cos(c + d*x))^2,x)

[Out]

int((1/cos(c + d*x))^(9/2)/(a + b/cos(c + d*x))^2, x)

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