3.57 \(\int \frac{(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=193 \[ -\frac{2 b \left (5 a^2-4 b^2\right ) \sqrt{e \sin (c+d x)}}{21 d e^5}-\frac{2 \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right ) (a+b \cos (c+d x))}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac{2 a \left (5 a^2-6 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 d e^4 \sqrt{e \sin (c+d x)}}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}} \]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(7*d*e*(e*Sin[c + d*x])^(7/2)) - (2*(a + b*Cos[c + d*x])*(a*b
+ (5*a^2 - 4*b^2)*Cos[c + d*x]))/(21*d*e^3*(e*Sin[c + d*x])^(3/2)) + (2*a*(5*a^2 - 6*b^2)*EllipticF[(c - Pi/2
+ d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*e^4*Sqrt[e*Sin[c + d*x]]) - (2*b*(5*a^2 - 4*b^2)*Sqrt[e*Sin[c + d*x]])
/(21*d*e^5)

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Rubi [A]  time = 0.273566, antiderivative size = 193, 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, 2861, 2669, 2642, 2641} \[ -\frac{2 b \left (5 a^2-4 b^2\right ) \sqrt{e \sin (c+d x)}}{21 d e^5}-\frac{2 \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right ) (a+b \cos (c+d x))}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac{2 a \left (5 a^2-6 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{21 d e^4 \sqrt{e \sin (c+d x)}}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(9/2),x]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(7*d*e*(e*Sin[c + d*x])^(7/2)) - (2*(a + b*Cos[c + d*x])*(a*b
+ (5*a^2 - 4*b^2)*Cos[c + d*x]))/(21*d*e^3*(e*Sin[c + d*x])^(3/2)) + (2*a*(5*a^2 - 6*b^2)*EllipticF[(c - Pi/2
+ d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*e^4*Sqrt[e*Sin[c + d*x]]) - (2*b*(5*a^2 - 4*b^2)*Sqrt[e*Sin[c + d*x]])
/(21*d*e^5)

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 2861

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*(d + c*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 - 1)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[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))^{9/2}} \, dx &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac{2 \int \frac{(a+b \cos (c+d x)) \left (-\frac{5 a^2}{2}+2 b^2-\frac{1}{2} a b \cos (c+d x)\right )}{(e \sin (c+d x))^{5/2}} \, dx}{7 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac{2 (a+b \cos (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \cos (c+d x)\right )}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac{4 \int \frac{\frac{1}{4} a \left (5 a^2-6 b^2\right )-\frac{1}{4} b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{21 e^4}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac{2 (a+b \cos (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \cos (c+d x)\right )}{21 d e^3 (e \sin (c+d x))^{3/2}}-\frac{2 b \left (5 a^2-4 b^2\right ) \sqrt{e \sin (c+d x)}}{21 d e^5}+\frac{\left (a \left (5 a^2-6 b^2\right )\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{21 e^4}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac{2 (a+b \cos (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \cos (c+d x)\right )}{21 d e^3 (e \sin (c+d x))^{3/2}}-\frac{2 b \left (5 a^2-4 b^2\right ) \sqrt{e \sin (c+d x)}}{21 d e^5}+\frac{\left (a \left (5 a^2-6 b^2\right ) \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{21 e^4 \sqrt{e \sin (c+d x)}}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac{2 (a+b \cos (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \cos (c+d x)\right )}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac{2 a \left (5 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)}}{21 d e^4 \sqrt{e \sin (c+d x)}}-\frac{2 b \left (5 a^2-4 b^2\right ) \sqrt{e \sin (c+d x)}}{21 d e^5}\\ \end{align*}

Mathematica [A]  time = 0.608406, size = 144, normalized size = 0.75 \[ -\frac{2 \csc ^4(c+d x) \sqrt{e \sin (c+d x)} \left (\frac{1}{4} \left (a \left (17 a^2+30 b^2\right ) \cos (c+d x)+36 a^2 b-5 a^3 \cos (3 (c+d x))+6 a b^2 \cos (3 (c+d x))+14 b^3 \cos (2 (c+d x))-2 b^3\right )+a \left (5 a^2-6 b^2\right ) \sin ^{\frac{7}{2}}(c+d x) F\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{21 d e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(9/2),x]

[Out]

(-2*Csc[c + d*x]^4*Sqrt[e*Sin[c + d*x]]*((36*a^2*b - 2*b^3 + a*(17*a^2 + 30*b^2)*Cos[c + d*x] + 14*b^3*Cos[2*(
c + d*x)] - 5*a^3*Cos[3*(c + d*x)] + 6*a*b^2*Cos[3*(c + d*x)])/4 + a*(5*a^2 - 6*b^2)*EllipticF[(-2*c + Pi - 2*
d*x)/4, 2]*Sin[c + d*x]^(7/2)))/(21*d*e^5)

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Maple [A]  time = 2.634, size = 241, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{\frac{2\,b \left ( 7\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+9\,{a}^{2}-4\,{b}^{2} \right ) }{21\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}}-{\frac{a}{21\,{e}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}\cos \left ( dx+c \right ) } \left ( \left ( -10\,{a}^{2}+12\,{b}^{2} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+ \left ( 16\,{a}^{2}+6\,{b}^{2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +5\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{9/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}-6\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{9/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2} \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))^(9/2),x)

[Out]

(-2/21*b/e/(e*sin(d*x+c))^(7/2)*(7*b^2*cos(d*x+c)^2+9*a^2-4*b^2)-1/21*a/e^4*((-10*a^2+12*b^2)*sin(d*x+c)*cos(d
*x+c)^4+(16*a^2+6*b^2)*cos(d*x+c)^2*sin(d*x+c)+5*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(9/2)*
EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-6*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(9/2)
*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2)/sin(d*x+c)^4/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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(9/2), 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^{5} \cos \left (d x + c\right )^{4} - 2 \, e^{5} \cos \left (d x + c\right )^{2} + e^{5}\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))^(9/2),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^5
*cos(d*x + c)^4 - 2*e^5*cos(d*x + c)^2 + e^5)*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))**(9/2),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{9}{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))^(9/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(9/2), x)