3.538 \(\int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=425 \[ -\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}+\frac {8 a b \sec ^3(c+d x)}{d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}+\frac {2 b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^4-114 a^2 b^2-15 b^4\right )-4 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^4}+\frac {\left (4 a^4-21 a^2 b^2-15 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

[Out]

2/3*b*sec(d*x+c)^3/(a^2-b^2)/d/(a+b*sin(d*x+c))^(3/2)+8*a*b*sec(d*x+c)^3/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)-
1/3*sec(d*x+c)^3*(b*(29*a^2+3*b^2)-a*(a^2+31*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^3/d-1/6*sec(d*x
+c)*(b*(a^4-114*a^2*b^2-15*b^4)-4*a*(a^4-6*a^2*b^2-27*b^4)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^4/d+2/
3*a*(a^4-6*a^2*b^2-27*b^4)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1
/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^4/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/
6*(4*a^4-21*a^2*b^2-15*b^4)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+
1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.88, antiderivative size = 425, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^3}+\frac {8 a b \sec ^3(c+d x)}{d \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}+\frac {2 b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (-114 a^2 b^2+a^4-15 b^4\right )-4 a \left (-6 a^2 b^2+a^4-27 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^4}+\frac {\left (-21 a^2 b^2+4 a^4-15 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (-6 a^2 b^2+a^4-27 b^4\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(2*b*Sec[c + d*x]^3)/(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^(3/2)) + (8*a*b*Sec[c + d*x]^3)/((a^2 - b^2)^2*d*Sq
rt[a + b*Sin[c + d*x]]) - (2*a*(a^4 - 6*a^2*b^2 - 27*b^4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(3*(a^2 - b^2)^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((4*a^4 - 21*a^2*b^2 - 15*b^4)*Ell
ipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(6*(a^2 - b^2)^3*d*Sqrt[a + b*Si
n[c + d*x]]) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(b*(29*a^2 + 3*b^2) - a*(a^2 + 31*b^2)*Sin[c + d*x]))/
(3*(a^2 - b^2)^3*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(b*(a^4 - 114*a^2*b^2 - 15*b^4) - 4*a*(a^4 - 6*a^
2*b^2 - 27*b^4)*Sin[c + d*x]))/(6*(a^2 - b^2)^4*d)

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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac {2 b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac {2 \int \frac {\sec ^4(c+d x) \left (-\frac {3 a}{2}+\frac {9}{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 ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \int \frac {\sec ^4(c+d x) \left (\frac {3}{4} \left (a^2+3 b^2\right )-21 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 ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {4 \int \frac {\sec ^2(c+d x) \left (-\frac {3}{8} \left (4 a^4-21 a^2 b^2-15 b^4\right )-\frac {9}{8} a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{9 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^4-114 a^2 b^2-15 b^4\right )-4 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}+\frac {4 \int \frac {-\frac {3}{16} b^2 \left (a^4-114 a^2 b^2-15 b^4\right )-\frac {3}{4} a b \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{9 \left (a^2-b^2\right )^4}\\ &=\frac {2 b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^4-114 a^2 b^2-15 b^4\right )-4 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}-\frac {\left (a \left (a^4-6 a^2 b^2-27 b^4\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^4}+\frac {\left (4 a^4-21 a^2 b^2-15 b^4\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{12 \left (a^2-b^2\right )^3}\\ &=\frac {2 b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^4-114 a^2 b^2-15 b^4\right )-4 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}-\frac {\left (a \left (a^4-6 a^2 b^2-27 b^4\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{3 \left (a^2-b^2\right )^4 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (\left (4 a^4-21 a^2 b^2-15 b^4\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}{12 \left (a^2-b^2\right )^3 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {2 b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac {8 a b \sec ^3(c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 a \left (a^4-6 a^2 b^2-27 b^4\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 )^4 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (4 a^4-21 a^2 b^2-15 b^4\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}}}{6 \left (a^2-b^2\right )^3 d \sqrt {a+b \sin (c+d x)}}-\frac {\sec ^3(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (29 a^2+3 b^2\right )-a \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^4-114 a^2 b^2-15 b^4\right )-4 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^4 d}\\ \end {align*}

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Mathematica [A]  time = 2.50, size = 341, normalized size = 0.80 \[ \frac {\frac {2 \left (a^2-b^2\right ) \sec ^3(c+d x) (a+b \sin (c+d x))^2 \left (a \left (a^2+3 b^2\right ) \sin (c+d x)-b \left (3 a^2+b^2\right )\right )+4 b^5 \left (a^2-b^2\right ) \cos (c+d x)+\sec (c+d x) (a+b \sin (c+d x))^2 \left (-a^4 b+54 a^2 b^3+4 a \left (a^4-6 a^2 b^2-11 b^4\right ) \sin (c+d x)+11 b^5\right )+64 a b^5 \cos (c+d x) (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {\left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2} \left (4 \left (a^5-6 a^3 b^2-27 a b^4\right ) E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\left (-4 a^5+4 a^4 b+21 a^3 b^2-21 a^2 b^3+15 a b^4-15 b^5\right ) F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )\right )}{(a-b)^4 (a+b)^2}}{6 d (a+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(((4*(a^5 - 6*a^3*b^2 - 27*a*b^4)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + (-4*a^5 + 4*a^4*b + 21*a^3
*b^2 - 21*a^2*b^3 + 15*a*b^4 - 15*b^5)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)])*((a + b*Sin[c + d*x])/
(a + b))^(3/2))/((a - b)^4*(a + b)^2) + (4*b^5*(a^2 - b^2)*Cos[c + d*x] + 64*a*b^5*Cos[c + d*x]*(a + b*Sin[c +
d*x]) + 2*(a^2 - b^2)*Sec[c + d*x]^3*(a + b*Sin[c + d*x])^2*(-(b*(3*a^2 + b^2)) + a*(a^2 + 3*b^2)*Sin[c + d*x
]) + Sec[c + d*x]*(a + b*Sin[c + d*x])^2*(-(a^4*b) + 54*a^2*b^3 + 11*b^5 + 4*a*(a^4 - 6*a^2*b^2 - 11*b^4)*Sin[
c + d*x]))/(a^2 - b^2)^4)/(6*d*(a + b*Sin[c + d*x])^(3/2))

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^4/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(b*sin(d*x + c) + a)^(5/2), x)

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maple [B]  time = 5.84, size = 2585, normalized size = 6.08 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sin(d*x+c))^(5/2),x)

[Out]

(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*(-1/4*(-a-3*b)/(a+b)^4/b/cos(d*x+c)^2/(a+b*sin(d*x+c))*(cos(d*x+c)^2*s
in(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*(EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-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)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^2-Elli
pticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(b/(a-b)*s
in(d*x+c)+1/(a-b)*a)^(1/2)*(-b/(a-b)*sin(d*x+c)-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)+4*a*b^4/(a+b)^3/(a-b)^3*(2*b*cos(d*x+c)^2/(a^2-b^2)/(-(-a-b*sin(d*x+c))*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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)
/(-(-a-b*sin(d*x+c))*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*b/(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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(
-(-a-b*sin(d*x+c))*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))))+1/4/(a+b)^2*(1/3/(a+b)*(-(-a-b*sin(d*x+c))*co
s(d*x+c)^2)^(1/2)/(sin(d*x+c)-1)^2-1/3*(-sin(d*x+c)^2*b-a*sin(d*x+c)-b*sin(d*x+c)-a)/(a+b)^2*(a+3*b)/((-a-b*si
n(d*x+c))*(sin(d*x+c)-1)*(1+sin(d*x+c)))^(1/2)+2*b^2/(3*a^2+6*a*b+3*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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*Ellip
ticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))-1/3*b*(a+3*b)/(a+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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*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^4/(a+b)^2/(a-b)^2*(2/3/b/(a^2-b^2)*(-(-a-b*sin(d*x+c))*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/(-(-a-b*sin(d*x+c))*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)*((-1-sin(d*x+c))*b/(a-b))^
(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*EllipticF(((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)*((-1-sin(d*x+c))*b/(a-
b))^(1/2)/(-(-a-b*sin(d*x+c))*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))))+1/4*(a-3*b)/(a-b)^3*(-(-sin(d*x+c)
^2*b-a*sin(d*x+c)+b*sin(d*x+c)+a)/(a-b)/((-a-b*sin(d*x+c))*(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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*
sin(d*x+c))*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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+
c))*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))))+1/4/(a-b)^2*(-1/3/(a-b)*(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1
/2)/(1+sin(d*x+c))^2-1/3*(-sin(d*x+c)^2*b-a*sin(d*x+c)+b*sin(d*x+c)+a)/(a-b)^2*(a-3*b)/((-a-b*sin(d*x+c))*(sin
(d*x+c)-1)*(1+sin(d*x+c)))^(1/2)+2*b^2/(3*a^2-6*a*b+3*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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin
(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))-1/3*b*(a-3*b)/(a-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)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*((-a/b-1)*Ell
ipticE(((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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sin(c + d*x))**(5/2), x)

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