3.667 \(\int \frac{1}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=391 \[ \frac{2 \left (16 a^2 b^2+a^4-16 b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (-13 a^2 b^2+a^4+8 b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}-\frac{8 b \left (-7 a^2 b^2+2 a^4+4 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(2*(a^4 + 16*a^2*b^2 - 16*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[S
ec[c + d*x]])/(3*a^4*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (8*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*EllipticE[(c +
 d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^4*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sq
rt[Sec[c + d*x]]) + (2*b^2*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (
4*b^2*(5*a^2 - 3*b^2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(
a^4 - 13*a^2*b^2 + 8*b^4)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.022, antiderivative size = 391, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3847, 4100, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (-13 a^2 b^2+a^4+8 b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}+\frac{2 \left (16 a^2 b^2+a^4-16 b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{8 b \left (-7 a^2 b^2+2 a^4+4 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(2*(a^4 + 16*a^2*b^2 - 16*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[S
ec[c + d*x]])/(3*a^4*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (8*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*EllipticE[(c +
 d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^4*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sq
rt[Sec[c + d*x]]) + (2*b^2*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (
4*b^2*(5*a^2 - 3*b^2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(
a^4 - 13*a^2*b^2 + 8*b^4)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[Sec[c + d*x]])

Rule 3847

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

Rule 4100

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*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
 b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

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{1}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{3 a^2}{2}+3 b^2+\frac{3}{2} a b \sec (c+d x)-2 b^2 \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} \left (a^4-13 a^2 b^2+8 b^4\right )-\frac{1}{2} a b \left (3 a^2-b^2\right ) \sec (c+d x)+b^2 \left (5 a^2-3 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}-\frac{8 \int \frac{\frac{3}{2} b \left (2 a^4-7 a^2 b^2+4 b^4\right )-\frac{3}{8} a \left (a^4+7 a^2 b^2-4 b^4\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{9 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}+\frac{\left (a^4+16 a^2 b^2-16 b^4\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac{\left (4 b \left (2 a^4-7 a^2 b^2+4 b^4\right )\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}+\frac{\left (\left (a^4+16 a^2 b^2-16 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}{3 a^4 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (4 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^4 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}+\frac{\left (\left (a^4+16 a^2 b^2-16 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}{3 a^4 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (4 b \left (2 a^4-7 a^2 b^2+4 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}{3 a^4 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \left (a^4+16 a^2 b^2-16 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)}}{3 a^4 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{8 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a^4 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{4 b^2 \left (5 a^2-3 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^4-13 a^2 b^2+8 b^4\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.63982, size = 257, normalized size = 0.66 \[ \frac{2 \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) \left (\frac{\left (\frac{a \cos (c+d x)+b}{a+b}\right )^{3/2} \left (\left (16 a^3 b^2-16 a^2 b^3-a^4 b+a^5-16 a b^4+16 b^5\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )-4 \left (-7 a^2 b^3+2 a^4 b+4 b^5\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{(a-b)^2}+\frac{a \sin (c+d x) \left (4 a b \left (-8 a^2 b^2+a^4+5 b^4\right ) \cos (c+d x)+\left (a^3-a b^2\right )^2 \cos (2 (c+d x))-25 a^2 b^4+a^6+16 b^6\right )}{2 \left (a^2-b^2\right )^2}\right )}{3 a^4 d (a+b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(5/2)*((((b + a*Cos[c + d*x])/(a + b))^(3/2)*(-4*(2*a^4*b - 7*a^2*b^3 + 4
*b^5)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] + (a^5 - a^4*b + 16*a^3*b^2 - 16*a^2*b^3 - 16*a*b^4 + 16*b^5)*Elli
pticF[(c + d*x)/2, (2*a)/(a + b)]))/(a - b)^2 + (a*(a^6 - 25*a^2*b^4 + 16*b^6 + 4*a*b*(a^4 - 8*a^2*b^2 + 5*b^4
)*Cos[c + d*x] + (a^3 - a*b^2)^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(2*(a^2 - b^2)^2)))/(3*a^4*d*(a + b*Sec[c + d
*x])^(5/2))

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Maple [B]  time = 0.337, size = 3614, normalized size = 9.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x)

[Out]

-2/3/d/a^4/(a-b)/(a+b)^2/((a-b)/(a+b))^(1/2)*(((a-b)/(a+b))^(1/2)*cos(d*x+c)^4*a^5*b-((a-b)/(a+b))^(1/2)*cos(d
*x+c)^4*a^4*b^2-((a-b)/(a+b))^(1/2)*cos(d*x+c)^4*a^3*b^3-a^4*b^2*((a-b)/(a+b))^(1/2)+7*a^3*b^3*((a-b)/(a+b))^(
1/2)+20*a^2*b^4*((a-b)/(a+b))^(1/2)-8*a*b^5*((a-b)/(a+b))^(1/2)-12*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^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)-16*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*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)+((a-b)/(a+b))^(1/2)*cos(d*x+c)
^4*a^6-((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^6+16*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^6+cos(d*x+c)^2*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^6-16*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
^6+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^6+28*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^4-16*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^5+1
0*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*b+25*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^2+4*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^3
-28*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^4-16*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^5-8*cos(d*x+c)^2*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
*b+28*cos(d*x+c)^2*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^3*b^3-16*cos(d*x+c)^2*sin(d*x+c)*Ell
ipticE((-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^5+9*cos(d*x+c)^2*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*b+16*cos(d*x+c)^2*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)*E
llipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*b^2-12*cos(d*x+c)^2*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^3-16*cos(d*x+c)^2*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^4-8*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*b-8*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^2+28*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^3*b^3-6*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^5*b-6*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^4*b^2+6*((a
-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^3*b^3+6*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b^4+7*((a-b)/(a+b))^(1/2)*cos(d*x
+c)^2*a^5*b-6*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^4*b^2-34*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^3*b^3+8*((a-b)/(a
+b))^(1/2)*cos(d*x+c)^2*a^2*b^4+24*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b^5-2*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^5
*b+14*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^4*b^2+22*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*b^3-34*((a-b)/(a+b))^(1/2)*
cos(d*x+c)*a^2*b^4-16*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^5-16*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*b^6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c)-8*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^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)+28*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^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)+EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^5*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)+9*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^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)+16*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^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)-16*b^6*(
(a-b)/(a+b))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1/cos(d*x+c))^(3/2)*cos(d*x+c)^2/sin(d*x+c)/(b+a*cos(
d*x+c))^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{5} + 3 \, a b^{2} \sec \left (d x + c\right )^{4} + 3 \, a^{2} b \sec \left (d x + c\right )^{3} + a^{3} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b^3*sec(d*x + c)^5 + 3*a*b^2*sec(d*x + c)^4 + 3*a^2*b*se
c(d*x + c)^3 + a^3*sec(d*x + c)^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(1/sec(d*x+c)**(3/2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2)), x)