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

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

[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]])

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Rubi [A]  time = 0.262706, antiderivative size = 219, normalized size of antiderivative = 1., 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 = {2693, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a \cos (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{4 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{4 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{2 \cos (c+d x)}{3 b d (a+b \sin (c+d x))^{3/2}} \]

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 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 2754

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]

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{\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*}

Mathematica [A]  time = 1.00898, size = 167, normalized size = 0.76 \[ \frac{2 b \cos (c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )+4 (a-b) (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-4 a (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{3 b^2 d (a-b) (a+b) (a+b \sin (c+d x))^{3/2}} \]

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]  time = 0.535, size = 864, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

2/3*(2*a*b^3*cos(d*x+c)^2*sin(d*x+c)-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)*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*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/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)
*a^3*b-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/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2)*a*b^3-2*(-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^4+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^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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \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(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \cos \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(cos(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*cos(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(cos(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{\cos \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(cos(d*x+c)^2/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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