3.6.25 \(\int \frac {\sec ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) [525]

Optimal. Leaf size=251 \[ \frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\left (a^2+3 b^2\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{\left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d} \]

[Out]

2*b*sec(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)-sec(d*x+c)*(4*a*b-(a^2+3*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(
1/2)/(a^2-b^2)^2/d+(a^2+3*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)^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2
)-a*(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)/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2773, 2945, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )^2}+\frac {2 b \sec (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}+\frac {a \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 )}{d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}-\frac {\left (a^2+3 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 )}{d \left (a^2-b^2\right )^2 \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])^(3/2),x]

[Out]

(2*b*Sec[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (a*EllipticF[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) -
(Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a*b - (a^2 + 3*b^2)*Sin[c + d*x]))/((a^2 - b^2)^2*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 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))^{3/2}} \, dx &=\frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (-\frac {a}{2}+\frac {3}{2} b \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac {2 \int \frac {-a b^2-\frac {1}{4} b \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac {a \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}-\frac {\left (a^2+3 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}-\frac {\left (\left (a^2+3 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}{2 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (a \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}{2 \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}\\ &=\frac {2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\left (a^2+3 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)}}{\left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {a 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}}}{\left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]
time = 1.72, size = 205, normalized size = 0.82 \begin {gather*} \frac {\left (a^3+a^2 b+3 a b^2+3 b^3\right ) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-a \left (a^2-b^2\right ) F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}-\frac {1}{2} \sec (c+d x) \left (3 a^2 b+b^3+b \left (a^2+3 b^2\right ) \cos (2 (c+d x))-2 a \left (a^2-b^2\right ) \sin (c+d x)\right )}{(a-b)^2 (a+b)^2 d \sqrt {a+b \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3 + a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a
+ b)] - a*(a^2 - b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - (Se
c[c + d*x]*(3*a^2*b + b^3 + b*(a^2 + 3*b^2)*Cos[2*(c + d*x)] - 2*a*(a^2 - b^2)*Sin[c + d*x]))/2)/((a - b)^2*(a
+ b)^2*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1063\) vs. \(2(305)=610\).
time = 3.38, size = 1064, normalized size = 4.24

method	result	size
default	\(-\frac {\sqrt {b \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+a \left (\cos ^{2}\left (d x +c \right )\right )}\, \left (\sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, a^{3} b +3 \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, a^{2} b^{2}-\sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, a \,b^{3}-3 \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, b^{4}-\sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, a^{4}-2 \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, a^{2} b^{2}+3 \sqrt {-\frac {b \sin \left (d x +c \right )}{a +b}+\frac {b}{a +b}}\, \EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, b^{4}+a^{2} b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+3 b^{4} \left (\cos ^{2}\left (d x +c \right )\right )-a^{3} b \sin \left (d x +c \right )+a \,b^{3} \sin \left (d x +c \right )+a^{2} b^{2}-b^{4}\right )}{b \left (a^{2}-b^{2}\right ) \left (a -b \right ) \left (a +b \right ) \sqrt {-\left (a +b \sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\)	\(1064\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/b*(b*cos(d*x+c)^2*sin(d*x+c)+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)
+1/(a-b)*a)^(1/2)*EllipticF((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)*a^3*b+3*(-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)*EllipticF((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)*a^2*b^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)*EllipticF((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)*a*b^3-3*(-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)*EllipticF((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^4-(-b/(a+b)*sin(d*x+c)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^4-2*(-b/(a+b)*sin(d*
x+c)+b/(a+b))^(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)+1/
(a-b)*a)^(1/2)*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^2*b^2+3*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*(-b/(a-b)*sin(d*x+
c)-b/(a-b))^(1/2)*b^4+a^2*b^2*cos(d*x+c)^2+3*b^4*cos(d*x+c)^2-a^3*b*sin(d*x+c)+a*b^3*sin(d*x+c)+a^2*b^2-b^4)/(
a^2-b^2)/(a-b)/(a+b)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+sin(d*x+c)))^(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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 685, normalized size = 2.73 \begin {gather*} \frac {2 \, {\left (\sqrt {2} {\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (a^{4} - 3 \, a^{2} b^{2}\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 ) + 2 \, {\left (\sqrt {2} {\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (a^{4} - 3 \, a^{2} b^{2}\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 (-i \, a^{2} b^{2} - 3 i \, b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{3} b - 3 i \, a b^{3}\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 (i \, a^{2} b^{2} + 3 i \, b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (i \, a^{3} b + 3 i \, a b^{3}\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 (a^{2} b^{2} - b^{4} + {\left (a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{6 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\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))^(3/2),x, algorithm="fricas")

[Out]

1/6*(2*(sqrt(2)*(a^3*b - 3*a*b^3)*cos(d*x + c)*sin(d*x + c) + sqrt(2)*(a^4 - 3*a^2*b^2)*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) + 2*(sqrt(2)*(a^3*b - 3*a*b^3)*cos(d*x + c)*sin(d*x + c) + sqrt(2)*(a^4 - 3*a^2*b^2)
*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)*(-I*a^2*b^2 - 3*I*b^4)*cos(d*x + c)*sin(d*x +
c) + sqrt(2)*(-I*a^3*b - 3*I*a*b^3)*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)) - 3*(sqrt(2)*(I*a^2*b^2 + 3*I*b^4)*cos(d*x + c)*sin(d*x + c)
+ sqrt(2)*(I*a^3*b + 3*I*a*b^3)*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*(a^2*b^2 - b^4 + (a^2*b^2 + 3*b^4)*cos(d*x + c)^2 - (a^3
*b - a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)*sin(d*x + c) +
(a^5*b - 2*a^3*b^3 + a*b^5)*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 {3}{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))**(3/2),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*sin(c + d*x))**(3/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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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 )}^{3/2}} \,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))^(3/2)),x)

[Out]

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

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