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

Optimal. Leaf size=360 \[ -\frac{8 b \left (a^2+4 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{5 a^4 d \sqrt{a+b \sec (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}+\frac{2 \left (8 a^2 b^2+3 a^4-16 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{5 a^4 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.966353, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3847, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 a^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)}}-\frac{8 b \left (a^2+4 b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{5 a^4 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 a^2 b^2+3 a^4-16 b^4\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{5 a^4 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*(a^2 + 4*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]
])/(5*a^4*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^4 + 8*a^2*b^2 - 16*b^4)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*
Sqrt[a + b*Sec[c + d*x]])/(5*a^4*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*b^2
*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) + (2*(a^2 - 6*b^2)*Sqrt[a + b*Sec
[c + d*x]]*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)) - (2*b*(3*a^2 - 8*b^2)*Sqrt[a + b*Sec[c + d*
x]]*Sin[c + d*x])/(5*a^3*(a^2 - b^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 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{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{2 \int \frac{-\frac{a^2}{2}+3 b^2+\frac{1}{2} a b \sec (c+d x)-2 b^2 \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{-\frac{3}{4} b \left (3 a^2-8 b^2\right )+\frac{1}{4} a \left (3 a^2+2 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (a^2-6 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}-\frac{8 \int \frac{-\frac{3}{8} \left (3 a^4+8 a^2 b^2-16 b^4\right )+\frac{3}{8} a b \left (a^2+4 b^2\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}-\frac{\left (4 b \left (a^2+4 b^2\right )\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 a^4}+\frac{\left (3 a^4+8 a^2 b^2-16 b^4\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{5 a^4 \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}-\frac{\left (4 b \left (a^2+4 b^2\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}{5 a^4 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^4+8 a^2 b^2-16 b^4\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{5 a^4 \left (a^2-b^2\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}-\frac{\left (4 b \left (a^2+4 b^2\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}{5 a^4 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^4+8 a^2 b^2-16 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}{5 a^4 \left (a^2-b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=-\frac{8 b \left (a^2+4 b^2\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)}}{5 a^4 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (3 a^4+8 a^2 b^2-16 b^4\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{5 a^4 \left (a^2-b^2\right ) d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2-6 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (3 a^2-8 b^2\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 a^3 \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.29537, size = 250, normalized size = 0.69 \[ \frac{\sqrt{\sec (c+d x)} \left (-16 b \left (3 a^2 b^2+a^4-4 b^4\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+2 a \sin (c+d x) \left (-4 a b \left (a^2-b^2\right ) \cos (c+d x)+\left (a^4-a^2 b^2\right ) \cos (2 (c+d x))-7 a^2 b^2+a^4+16 b^4\right )+4 \left (8 a^3 b^2+8 a^2 b^3+3 a^4 b+3 a^5-16 a b^4-16 b^5\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )\right )}{10 a^4 d (a-b) (a+b) \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sec[c + d*x]]*(4*(3*a^5 + 3*a^4*b + 8*a^3*b^2 + 8*a^2*b^3 - 16*a*b^4 - 16*b^5)*Sqrt[(b + a*Cos[c + d*x])
/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - 16*b*(a^4 + 3*a^2*b^2 - 4*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a
+ b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(a^4 - 7*a^2*b^2 + 16*b^4 - 4*a*b*(a^2 - b^2)*Cos[c + d*x] +
(a^4 - a^2*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(10*a^4*(a - b)*(a + b)*d*Sqrt[a + b*Sec[c + d*x]])

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Maple [B]  time = 0.305, size = 1861, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d/a^4/(a+b)/((a-b)/(a+b))^(1/2)*(8*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2-2*cos(d*x+c)^3*((a-b)/(a+b))^
(1/2)*a^3*b-3*a^3*b*((a-b)/(a+b))^(1/2)-8*a*b^3*((a-b)/(a+b))^(1/2)-3*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+8*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^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)-4*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*(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)-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^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*E
llipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*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*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin
(d*x+c),(-(a+b)/(a-b))^(1/2))*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)-3*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^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*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^4+16*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^4-3*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4+cos(d*x
+c)^4*((a-b)/(a+b))^(1/2)*a^4+2*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^4-16*b^4*((a-b)/(a+b))^(1/2)+cos(d*x+c)^4*(
(a-b)/(a+b))^(1/2)*a^3*b-2*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b^2+2*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b+8
*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^3+2*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b-6*cos(d*x+c)*((a-b)/(a+b))^(1/2
)*a^2*b^2+3*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)*cos(d*x+c)*sin(d*x+c)*a^4+8*cos(d*x+c)*sin(d*x+c)*Ellipt
icE((-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^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-12*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)*Ellipt
icF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2-16*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^3+3*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*a^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))*((b+a*
cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/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{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), 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^{2} \sec \left (d x + c\right )^{5} + 2 \, a b \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b^2*sec(d*x + c)^5 + 2*a*b*sec(d*x + c)^4 + a^2*sec(d*x
+ c)^3), 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)**(5/2)/(a+b*sec(d*x+c))**(3/2),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{3}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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