3.646 \(\int \frac{(a+b \sec (c+d x))^{5/2}}{\sec ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=363 \[ \frac{4 b \left (-62 a^2 b^2+57 a^4+5 b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{315 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (49 a^2+75 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{315 a d \sqrt{\sec (c+d x)}}+\frac{2 \left (279 a^2 b^2+147 a^4-10 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{315 a^2 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a^2 \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{63 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(4*b*(57*a^4 - 62*a^2*b^2 + 5*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sq
rt[Sec[c + d*x]])/(315*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(147*a^4 + 279*a^2*b^2 - 10*b^4)*EllipticE[(c + d*
x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(315*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]
]) + (2*a^2*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (38*a*b*Sqrt[a + b*Sec[c + d*x]]
*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*(49*a^2 + 75*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(315*d*
Sec[c + d*x]^(3/2)) + (2*b*(163*a^2 + 5*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(315*a*d*Sqrt[Sec[c + d*x]
])

________________________________________________________________________________________

Rubi [A]  time = 1.25392, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3841, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (49 a^2+75 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{315 a d \sqrt{\sec (c+d x)}}+\frac{4 b \left (-62 a^2 b^2+57 a^4+5 b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{315 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (279 a^2 b^2+147 a^4-10 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{315 a^2 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a^2 \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{63 d \sec ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(9/2),x]

[Out]

(4*b*(57*a^4 - 62*a^2*b^2 + 5*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sq
rt[Sec[c + d*x]])/(315*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(147*a^4 + 279*a^2*b^2 - 10*b^4)*EllipticE[(c + d*
x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(315*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]
]) + (2*a^2*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + (38*a*b*Sqrt[a + b*Sec[c + d*x]]
*Sin[c + d*x])/(63*d*Sec[c + d*x]^(5/2)) + (2*(49*a^2 + 75*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(315*d*
Sec[c + d*x]^(3/2)) + (2*b*(163*a^2 + 5*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(315*a*d*Sqrt[Sec[c + d*x]
])

Rule 3841

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

Rule 4104

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]

Rule 4035

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

Rule 3856

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

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

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

Rule 3858

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^{5/2}}{\sec ^{\frac{9}{2}}(c+d x)} \, dx &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2}{9} \int \frac{\frac{19 a^2 b}{2}+\frac{1}{2} a \left (7 a^2+27 b^2\right ) \sec (c+d x)+\frac{3}{2} b \left (2 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{7}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{4 \int \frac{-\frac{1}{4} a^2 \left (49 a^2+75 b^2\right )-\frac{1}{4} a b \left (137 a^2+63 b^2\right ) \sec (c+d x)-19 a^2 b^2 \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{63 a}\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{\frac{3}{8} a^2 b \left (163 a^2+5 b^2\right )+\frac{1}{8} a^3 \left (147 a^2+605 b^2\right ) \sec (c+d x)+\frac{1}{4} a^2 b \left (49 a^2+75 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{315 a^2}\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 a d \sqrt{\sec (c+d x)}}-\frac{16 \int \frac{-\frac{3}{16} a^2 \left (147 a^4+279 a^2 b^2-10 b^4\right )-\frac{3}{16} a^3 b \left (261 a^2+155 b^2\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{945 a^3}\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 a d \sqrt{\sec (c+d x)}}+\frac{\left (147 a^4+279 a^2 b^2-10 b^4\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{315 a^2}+\frac{\left (2 b \left (57 a^4-62 a^2 b^2+5 b^4\right )\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{315 a^2}\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 a d \sqrt{\sec (c+d x)}}+\frac{\left (2 b \left (57 a^4-62 a^2 b^2+5 b^4\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{315 a^2 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (147 a^4+279 a^2 b^2-10 b^4\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{315 a^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 a d \sqrt{\sec (c+d x)}}+\frac{\left (2 b \left (57 a^4-62 a^2 b^2+5 b^4\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{315 a^2 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (147 a^4+279 a^2 b^2-10 b^4\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{315 a^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{4 b \left (57 a^4-62 a^2 b^2+5 b^4\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{315 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (147 a^4+279 a^2 b^2-10 b^4\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{315 a^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 a^2 \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{9 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{38 a b \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{63 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (49 a^2+75 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 b \left (163 a^2+5 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{315 a d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.57798, size = 286, normalized size = 0.79 \[ \frac{(a+b \sec (c+d x))^{5/2} \left (32 b \left (-62 a^2 b^2+57 a^4+5 b^4\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+2 a \sin (c+d x) \left (4 a b \left (619 a^2+160 b^2\right ) \cos (c+d x)+8 \left (85 a^2 b^2+42 a^4\right ) \cos (2 (c+d x))+1984 a^2 b^2+260 a^3 b \cos (3 (c+d x))+35 a^4 \cos (4 (c+d x))+301 a^4+40 b^4\right )+16 \left (279 a^3 b^2+279 a^2 b^3+147 a^4 b+147 a^5-10 a b^4-10 b^5\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{2520 a^2 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(9/2),x]

[Out]

((a + b*Sec[c + d*x])^(5/2)*(16*(147*a^5 + 147*a^4*b + 279*a^3*b^2 + 279*a^2*b^3 - 10*a*b^4 - 10*b^5)*Sqrt[(b
+ a*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a + b)] + 32*b*(57*a^4 - 62*a^2*b^2 + 5*b^4)*Sqrt[(b
+ a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(301*a^4 + 1984*a^2*b^2 + 40*b^4 + 4*a*
b*(619*a^2 + 160*b^2)*Cos[c + d*x] + 8*(42*a^4 + 85*a^2*b^2)*Cos[2*(c + d*x)] + 260*a^3*b*Cos[3*(c + d*x)] + 3
5*a^4*Cos[4*(c + d*x)])*Sin[c + d*x]))/(2520*a^2*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.465, size = 2788, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(9/2),x)

[Out]

-2/315/d/a^2/((a-b)/(a+b))^(1/2)*(-147*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))
^(1/2))*a^4*b*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-147*a^4*b*((
a-b)/(a+b))^(1/2)-163*a^3*b^2*((a-b)/(a+b))^(1/2)-279*a^2*b^3*((a-b)/(a+b))^(1/2)-5*a*b^4*((a-b)/(a+b))^(1/2)+
147*cos(d*x+c)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a
+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^5+10*cos(d*x+c)*sin(d*x+c)*EllipticE((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1
/2)*(1/(cos(d*x+c)+1))^(1/2)*b^5-147*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/
(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^5+279*E
llipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b^2*(1/(a+b)*(b+a*cos(d*x+c)
)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-279*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/
sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)-10*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^4*(1/(a+b
)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+261*EllipticF((-1+cos(d*x+c))*((a
-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*b*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(co
s(d*x+c)+1))^(1/2)*sin(d*x+c)-279*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2
))*a^3*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+155*EllipticF((
-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+10*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*a*b^4*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)-147*cos(d*x+c)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^4*b+279*cos(d*x+c)*sin(d*x+c)*Elliptic
E((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^3*b^2-279*cos(d*x+c)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1
/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*
a^2*b^3-10*cos(d*x+c)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2)
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b^4+261*cos(d*x+c)*sin(d*x+c)*(1/
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*b-279*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2
))*a^3*b^2+155*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*
EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^3+10*cos(d*x+c)*sin(d*x+c
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a
+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^4+10*b^5*((a-b)/(a+b))^(1/2)+147*EllipticE((-1+cos(d*x+c))*((a
-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c)+10*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*
b^5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-147*EllipticF((-1+cos(
d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/
2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+35*cos(d*x+c)^6*((a-b)/(a+b))^(1/2)*a^5+14*cos(d*x+c)^4*((a-b)/(a+b))^(
1/2)*a^5+98*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^5-147*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^5-10*cos(d*x+c)*((a-b)/(
a+b))^(1/2)*b^5+170*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b^2+82*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^4*b+80*cos(
d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b^3+272*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b^2-5*cos(d*x+c)^2*((a-b)/(a+b))
^(1/2)*a*b^4-65*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4*b-279*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b^2+199*cos(d*x+c)
*((a-b)/(a+b))^(1/2)*a^2*b^3+10*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^4+130*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^4*
b)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^5*(1/cos(d*x+c))^(9/2)/sin(d*x+c)/(b+a*cos(d*x+c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{9}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(5/2)/sec(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)/sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(9/2), x)