3.6.36 \(\int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [536]

Optimal. Leaf size=219 \[ -\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {4 a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 b^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 b^2 d \sqrt {a+b \sin (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 219, 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 = {2772, 2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {4 a \cos (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}+\frac {4 a \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 b^2 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \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 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2772

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[g^2*((p - 1)/(b*(m + 1))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 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 2833

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx &=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}-\frac {2 \int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 b}\\ &=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {4 \int \frac {\frac {b}{2}+\frac {1}{2} a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}-\frac {2 \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{3 b^2}+\frac {(2 a) \int \sqrt {a+b \sin (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {\left (2 a \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{3 b^2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (2 \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}{3 b^2 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}}+\frac {4 a \cos (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt {a+b \sin (c+d x)}}+\frac {4 a 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 b^2 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 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 b^2 d \sqrt {a+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.05, size = 167, normalized size = 0.76 \begin {gather*} \frac {-4 a (a+b)^2 E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+4 (a-b) (a+b)^2 F\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}+2 b \cos (c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{3 (a-b) b^2 (a+b) d (a+b \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*(a + b)^2*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + 4*(a -
b)*(a + b)^2*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + 2*b*Cos[c
+ d*x]*(a^2 + b^2 + 2*a*b*Sin[c + d*x]))/(3*(a - b)*b^2*(a + b)*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. \(863\) vs. \(2(265)=530\).
time = 2.64, size = 864, normalized size = 3.95

method	result	size
default	\(\frac {\frac {4 a \,b^{3} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}-\frac {4 \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}}\, \sqrt {-\frac {b \sin \left (d x +c \right )}{a -b}-\frac {b}{a -b}}\, b \left (\EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}-\EllipticE \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}-\EllipticF \left (\sqrt {\frac {b \sin \left (d x +c \right )}{a -b}+\frac {a}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} 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 ) b^{3}\right ) \sin \left (d x +c \right )}{3}+\frac {2 \left (a^{2} b^{2}+b^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )}{3}+\frac {4 \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}-\frac {4 \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}-\frac {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}}{3}+\frac {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^{2} b^{2}}{3}}{\left (a^{2}-b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} b^{3} \cos \left (d x +c \right ) d}\)	\(864\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(2*a*b^3*cos(d*x+c)^2*sin(d*x+c)-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/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*b*(EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a
^3-EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^2-EllipticF((b/(a-b)*sin(d*x+c)+1/(
a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b+EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*b
^3)*sin(d*x+c)+(a^2*b^2+b^4)*cos(d*x+c)^2+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-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^3-2*(-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^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)/(a^2-b^2)/(a+b*sin(d*x+c))^(3/2)/b^3/cos(d*x+c)/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(cos(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(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 = 0.14, size = 711, normalized size = 3.25 \begin {gather*} -\frac {2 \, {\left ({\left (\sqrt {2} {\left (2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right ) - \sqrt {2} {\left (2 \, a^{4} - a^{2} b^{2} - 3 \, b^{4}\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 (2 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right ) - \sqrt {2} {\left (2 \, a^{4} - a^{2} b^{2} - 3 \, b^{4}\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 (i \, \sqrt {2} a b^{3} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{2} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{3} b - i \, a b^{3}\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 (-i \, \sqrt {2} a b^{3} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{2} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (i \, a^{3} b + i \, a b^{3}\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 (2 \, a b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}\right )}}{9 \, {\left ({\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{4} - a b^{6}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{3} - b^{7}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*((sqrt(2)*(2*a^2*b^2 - 3*b^4)*cos(d*x + c)^2 - 2*sqrt(2)*(2*a^3*b - 3*a*b^3)*sin(d*x + c) - sqrt(2)*(2*a^
4 - a^2*b^2 - 3*b^4))*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) + (sqrt(2)*(2*a^2*b^2 - 3*b^4)*cos(d*x + c)^2 - 2*sqrt
(2)*(2*a^3*b - 3*a*b^3)*sin(d*x + c) - sqrt(2)*(2*a^4 - a^2*b^2 - 3*b^4))*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*(I*sqrt(2)*a*b^3*cos(d*x + c)^2 - 2*I*sqrt(2)*a^2*b^2*sin(d*x + c) + sqrt(2)*(-I*a^3*b - I*a*b^3))*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*(-I
*sqrt(2)*a*b^3*cos(d*x + c)^2 + 2*I*sqrt(2)*a^2*b^2*sin(d*x + c) + sqrt(2)*(I*a^3*b + I*a*b^3))*sqrt(-I*b)*wei
erstrassZeta(-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*(2*a*b^
3*cos(d*x + c)*sin(d*x + c) + (a^2*b^2 + b^4)*cos(d*x + c))*sqrt(b*sin(d*x + c) + a))/((a^2*b^5 - b^7)*d*cos(d
*x + c)^2 - 2*(a^3*b^4 - a*b^6)*d*sin(d*x + c) - (a^4*b^3 - b^7)*d)

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

[Out]

Integral(cos(c + d*x)**2/(a + b*sin(c + d*x))**(5/2), 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(cos(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),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 {{\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(cos(c + d*x)^2/(a + b*sin(c + d*x))^(5/2),x)

[Out]

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

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