3.6.37 \(\int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [537]

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

[Out]

2/3*b*sec(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^(3/2)+16/3*a*b*sec(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^(1/2)-1
/3*sec(d*x+c)*(b*(27*a^2+5*b^2)-a*(3*a^2+29*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)^3/d+1/3*a*(3*a^2
+29*b^2)*(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)^3/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-1/3*(3*a^2+5*b^2)*(s
in(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)^2/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.39, antiderivative size = 325, normalized size of antiderivative = 1.00, 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 = {2773, 2943, 2945, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\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}}} \end {gather*}

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 2732

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

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2773

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] /;
FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2831

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 2943

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)*S
imp[(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 2945

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 ^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*}

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Mathematica [A]
time = 1.92, size = 241, normalized size = 0.74 \begin {gather*} \frac {\frac {\left (\left (3 a^3+29 a b^2\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )+\left (-3 a^3+3 a^2 b-5 a b^2+5 b^3\right ) F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right )\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}}{(a-b)^3 (a+b)}-\frac {2 b^3 \left (a^2-b^2\right ) \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 \left (3 a^2 b+b^3-a \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^3}}{3 d (a+b \sin (c+d x))^{3/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(1652\) vs. \(2(367)=734\).
time = 12.44, size = 1653, normalized size = 5.09

method	result	size
default	\(\text {Expression too large to display}\)	\(1653\)

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,method=_RETURNVERBOSE)

[Out]

(-(-a-b*sin(d*x+c))*cos(d*x+c)^2)^(1/2)*(1/2/(a+b)^3/b/cos(d*x+c)^2/(a+b*sin(d*x+c))*(b*cos(d*x+c)^2*sin(d*x+c
)+a*cos(d*x+c)^2)^(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-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)*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)-b^2/(a-b)/(a+b)*(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))*co
s(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))))-2*a*b^2/(a+b)^2/(a-b)^2*(2*b*cos(d*x+c)^2/(a^2-b^2)/(-(-a-b*si
n(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))))+1/2/(a-b)^
2*(-(-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)))))/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.18, size = 993, normalized size = 3.06 \begin {gather*} \frac {{\left (\sqrt {2} {\left (6 \, a^{4} b^{2} - 23 \, a^{2} b^{4} - 15 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (6 \, a^{5} b - 23 \, a^{3} b^{3} - 15 \, a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt {2} {\left (6 \, a^{6} - 17 \, a^{4} b^{2} - 38 \, a^{2} b^{4} - 15 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) + {\left (\sqrt {2} {\left (6 \, a^{4} b^{2} - 23 \, a^{2} b^{4} - 15 \, b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (6 \, a^{5} b - 23 \, a^{3} b^{3} - 15 \, a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt {2} {\left (6 \, a^{6} - 17 \, a^{4} b^{2} - 38 \, a^{2} b^{4} - 15 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sqrt {-i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right ) - 3 \, {\left (\sqrt {2} {\left (-3 i \, a^{3} b^{3} - 29 i \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (3 i \, a^{4} b^{2} + 29 i \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (3 i \, a^{5} b + 32 i \, a^{3} b^{3} + 29 i \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) - 3 \, {\left (\sqrt {2} {\left (3 i \, a^{3} b^{3} + 29 i \, a b^{5}\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-3 i \, a^{4} b^{2} - 29 i \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, a^{5} b - 32 i \, a^{3} b^{3} - 29 i \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) + 6 \, {\left (3 \, a^{4} b^{2} - 6 \, a^{2} b^{4} + 3 \, b^{6} + {\left (6 \, a^{4} b^{2} + 31 \, a^{2} b^{4} - 5 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, a^{5} b - 6 \, a^{3} b^{3} + 3 \, a b^{5} - {\left (3 \, a^{3} b^{3} + 29 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{18 \, {\left ({\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{8} b - 2 \, a^{6} b^{3} + 2 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )\right )}} \end {gather*}

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]

1/18*((sqrt(2)*(6*a^4*b^2 - 23*a^2*b^4 - 15*b^6)*cos(d*x + c)^3 - 2*sqrt(2)*(6*a^5*b - 23*a^3*b^3 - 15*a*b^5)*
cos(d*x + c)*sin(d*x + c) - sqrt(2)*(6*a^6 - 17*a^4*b^2 - 38*a^2*b^4 - 15*b^6)*cos(d*x + c))*sqrt(I*b)*weierst
rassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x
+ c) - 2*I*a)/b) + (sqrt(2)*(6*a^4*b^2 - 23*a^2*b^4 - 15*b^6)*cos(d*x + c)^3 - 2*sqrt(2)*(6*a^5*b - 23*a^3*b^3
- 15*a*b^5)*cos(d*x + c)*sin(d*x + c) - sqrt(2)*(6*a^6 - 17*a^4*b^2 - 38*a^2*b^4 - 15*b^6)*cos(d*x + c))*sqrt
(-I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) +
3*I*b*sin(d*x + c) + 2*I*a)/b) - 3*(sqrt(2)*(-3*I*a^3*b^3 - 29*I*a*b^5)*cos(d*x + c)^3 + 2*sqrt(2)*(3*I*a^4*b
^2 + 29*I*a^2*b^4)*cos(d*x + c)*sin(d*x + c) + sqrt(2)*(3*I*a^5*b + 32*I*a^3*b^3 + 29*I*a*b^5)*cos(d*x + c))*s
qrt(I*b)*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(
4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) -
3*(sqrt(2)*(3*I*a^3*b^3 + 29*I*a*b^5)*cos(d*x + c)^3 + 2*sqrt(2)*(-3*I*a^4*b^2 - 29*I*a^2*b^4)*cos(d*x + c)*s
in(d*x + c) + sqrt(2)*(-3*I*a^5*b - 32*I*a^3*b^3 - 29*I*a*b^5)*cos(d*x + c))*sqrt(-I*b)*weierstrassZeta(-4/3*(
4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*
I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) + 6*(3*a^4*b^2 - 6*a^2*b^4 + 3
*b^6 + (6*a^4*b^2 + 31*a^2*b^4 - 5*b^6)*cos(d*x + c)^2 - (3*a^5*b - 6*a^3*b^3 + 3*a*b^5 - (3*a^3*b^3 + 29*a*b^
5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*cos(d*x
+ c)^3 - 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*cos(d*x + c)*sin(d*x + c) - (a^8*b - 2*a^6*b^3 + 2*a^2*
b^7 - b^9)*d*cos(d*x + c))

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

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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