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

Optimal. Leaf size=325 \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}+\frac{16 a b \sec (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \sec (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac{\left (3 a^2+5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{a \left (3 a^2+29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

(2*b*Sec[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(3/2)) + (16*a*b*Sec[c + d*x])/(3*(a^2 - b^2)^2*d*Sqr
t[a + b*Sin[c + d*x]]) - (a*(3*a^2 + 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d
*x]])/(3*(a^2 - b^2)^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((3*a^2 + 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2,
(2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x
]*Sqrt[a + b*Sin[c + d*x]]*(b*(27*a^2 + 5*b^2) - a*(3*a^2 + 29*b^2)*Sin[c + d*x]))/(3*(a^2 - b^2)^3*d)

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Rubi [A]  time = 0.609488, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2694, 2864, 2866, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}+\frac{16 a b \sec (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \sec (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}+\frac{\left (3 a^2+5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{a \left (3 a^2+29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sec[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(3/2)) + (16*a*b*Sec[c + d*x])/(3*(a^2 - b^2)^2*d*Sqr
t[a + b*Sin[c + d*x]]) - (a*(3*a^2 + 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d
*x]])/(3*(a^2 - b^2)^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((3*a^2 + 5*b^2)*EllipticF[(c - Pi/2 + d*x)/2,
(2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*(a^2 - b^2)^2*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x
]*Sqrt[a + b*Sin[c + d*x]]*(b*(27*a^2 + 5*b^2) - a*(3*a^2 + 29*b^2)*Sin[c + d*x]))/(3*(a^2 - b^2)^3*d)

Rule 2694

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m +
1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; F
reeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2864

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a
^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*Sim
p[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[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 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]

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]

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (-\frac{3 a}{2}+\frac{5}{2} b \sin (c+d x)\right )}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{16 a b \sec (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \int \frac{\sec ^2(c+d x) \left (\frac{1}{4} \left (3 a^2+5 b^2\right )-6 a b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{16 a b \sec (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{4 \int \frac{\frac{1}{8} b^2 \left (27 a^2+5 b^2\right )+\frac{1}{8} a b \left (3 a^2+29 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{16 a b \sec (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac{\left (3 a^2+5 b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{6 \left (a^2-b^2\right )^2}-\frac{\left (a \left (3 a^2+29 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{16 a b \sec (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac{\left (a \left (3 a^2+29 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{6 \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (\left (3 a^2+5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{6 \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \sec (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{16 a b \sec (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{a \left (3 a^2+29 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{3 \left (a^2-b^2\right )^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (3 a^2+5 b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (27 a^2+5 b^2\right )-a \left (3 a^2+29 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 1.84628, size = 241, normalized size = 0.74 \[ \frac{\frac{\left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} \left (\left (3 a^2 b-3 a^3-5 a b^2+5 b^3\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (3 a^3+29 a b^2\right ) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )}{(a-b)^3 (a+b)}-\frac{2 b^3 \left (a^2-b^2\right ) \cos (c+d x)+3 \sec (c+d x) (a+b \sin (c+d x))^2 \left (-a \left (a^2+3 b^2\right ) \sin (c+d x)+3 a^2 b+b^3\right )+20 a b^3 \cos (c+d x) (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}}{3 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((((3*a^3 + 29*a*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + (-3*a^3 + 3*a^2*b - 5*a*b^2 + 5*b^3)*E
llipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)])*((a + b*Sin[c + d*x])/(a + b))^(3/2))/((a - b)^3*(a + b)) - (2
*b^3*(a^2 - b^2)*Cos[c + d*x] + 20*a*b^3*Cos[c + d*x]*(a + b*Sin[c + d*x]) + 3*Sec[c + d*x]*(a + b*Sin[c + d*x
])^2*(3*a^2*b + b^3 - a*(a^2 + 3*b^2)*Sin[c + d*x]))/(a^2 - b^2)^3)/(3*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [B]  time = 3.105, size = 1653, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*(-2*a*b^2/(a+b)^2/(a-b)^2*(2*b*cos(d*x+c)^2/(a^2-b^2)/(-(-b*sin(d*x+c)
-a)*cos(d*x+c)^2)^(1/2)+2*a/(a^2-b^2)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((
-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2)
,((a-b)/(a+b))^(1/2))+2/(a^2-b^2)*b*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-s
in(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b
))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))-b^2/(a+b)/(a-b)*
(2/3/(a^2-b^2)/b*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a/b)^2+8/3*b*cos(d*x+c)^2/(a^2-b^2)^2*a/(
-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)+2*(3*a^2+b^2)/(3*a^4-6*a^2*b^2+3*b^4)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^
(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*E
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+8/3*a*b/(a^2-b^2)^2*(a/b-1)*((a+b*sin(d*x+c))/(a-
b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/
2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^
(1/2),((a-b)/(a+b))^(1/2))))+1/2/(a+b)^3/b/cos(d*x+c)^2/(a+b*sin(d*x+c))*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+
c)^2)^(1/2)*((-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-
b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2-(-b/(a-b)*sin(d*x+c)-b/(a-
b))^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*EllipticE((b/(a-b)*sin(d*x+
c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b^2-b^2*cos(d*x+c)^2+a*b*sin(d*x+c)+b^2*sin(d*x+c)+a*b+b^2)+1/2/(a-b)
^2*(-(-sin(d*x+c)^2*b-a*sin(d*x+c)+b*sin(d*x+c)+a)/(a-b)/((-b*sin(d*x+c)-a)*(sin(d*x+c)-1)*(1+sin(d*x+c)))^(1/
2)-2*b/(2*a-2*b)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b
))^(1/2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))
-b/(a-b)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-sin(d*x+c)-1)*b/(a-b))^(1/2)
/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2
))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)))))/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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