3.55 \(\int \frac{(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a b \sqrt{e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}} \]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*(a^2 - 6*b^2)*EllipticF
[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]) - (2*b*(a^2 + 4*b^2)*Sqrt[e*Sin[c +
d*x]])/(3*d*e^3) - (2*a*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e^3)

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Rubi [A]  time = 0.259627, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2691, 2862, 2669, 2642, 2641} \[ -\frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 a b \sqrt{e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e^3}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*(a^2 - 6*b^2)*EllipticF
[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]) - (2*b*(a^2 + 4*b^2)*Sqrt[e*Sin[c +
d*x]])/(3*d*e^3) - (2*a*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e^3)

Rule 2691

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

Rule 2862

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

Rule 2669

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{5/2}} \, dx &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 \int \frac{(a+b \cos (c+d x)) \left (-\frac{a^2}{2}+2 b^2+\frac{3}{2} a b \cos (c+d x)\right )}{\sqrt{e \sin (c+d x)}} \, dx}{3 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{4 \int \frac{-\frac{3}{4} a \left (a^2-6 b^2\right )+\frac{3}{4} b \left (a^2+4 b^2\right ) \cos (c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{9 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{\left (a \left (a^2-6 b^2\right )\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{3 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{\left (a \left (a^2-6 b^2\right ) \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 e^2 \sqrt{e \sin (c+d x)}}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}+\frac{2 a \left (a^2-6 b^2\right ) F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{2 b \left (a^2+4 b^2\right ) \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{2 a b (a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}}{3 d e^3}\\ \end{align*}

Mathematica [A]  time = 0.826395, size = 102, normalized size = 0.6 \[ -\frac{2 a \left (a^2+3 b^2\right ) \cos (c+d x)+2 a \left (a^2-6 b^2\right ) \sin ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+6 a^2 b-3 b^3 \cos (2 (c+d x))+5 b^3}{3 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(5/2),x]

[Out]

-(6*a^2*b + 5*b^3 + 2*a*(a^2 + 3*b^2)*Cos[c + d*x] - 3*b^3*Cos[2*(c + d*x)] + 2*a*(a^2 - 6*b^2)*EllipticF[(-2*
c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(3/2))/(3*d*e*(e*Sin[c + d*x])^(3/2))

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Maple [A]  time = 2.434, size = 214, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{\frac{2\,b \left ( -3\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,{a}^{2}+4\,{b}^{2} \right ) }{3\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{a}{3\,{e}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) } \left ( \left ( 2\,{a}^{2}+6\,{b}^{2} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },{\frac{\sqrt{2}}{2}} \right ){a}^{2}-6\,{b}^{2}\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{5/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x)

[Out]

(-2/3*b/e/(e*sin(d*x+c))^(3/2)*(-3*b^2*cos(d*x+c)^2+3*a^2+4*b^2)-1/3*a/e^2*((2*a^2+6*b^2)*sin(d*x+c)*cos(d*x+c
)^2+(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a
^2-6*b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/
2)))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sqrt{e \sin \left (d x + c\right )}}{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sqrt(e*sin(d*x + c))/((e^
3*cos(d*x + c)^2 - e^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((a+b*cos(d*x+c))**3/(e*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{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(5/2), x)