3.48 \(\int \frac{(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d e^4 \sqrt{\sin (c+d x)}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.171197, antiderivative size = 165, 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, 2669, 2636, 2640, 2639} \[ -\frac{2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d e^4 \sqrt{\sin (c+d x)}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(7/2),x]

[Out]

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

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 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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2640

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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))^2}{(e \sin (c+d x))^{7/2}} \, dx &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac{2 \int \frac{-\frac{3 a^2}{2}+b^2-\frac{1}{2} a b \cos (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}+\frac{\left (3 a^2-2 b^2\right ) \int \frac{1}{(e \sin (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{\left (3 a^2-2 b^2\right ) \int \sqrt{e \sin (c+d x)} \, dx}{5 e^4}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{\left (\left (3 a^2-2 b^2\right ) \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 e^4 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{5 d e (e \sin (c+d x))^{5/2}}-\frac{2 a b}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) \cos (c+d x)}{5 d e^3 \sqrt{e \sin (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 d e^4 \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.489102, size = 109, normalized size = 0.66 \[ -\frac{\left (7 a^2+2 b^2\right ) \cos (c+d x)-4 \left (3 a^2-2 b^2\right ) \sin ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )-3 a^2 \cos (3 (c+d x))+8 a b+2 b^2 \cos (3 (c+d x))}{10 d e (e \sin (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2/(e*Sin[c + d*x])^(7/2),x]

[Out]

-(8*a*b + (7*a^2 + 2*b^2)*Cos[c + d*x] - 3*a^2*Cos[3*(c + d*x)] + 2*b^2*Cos[3*(c + d*x)] - 4*(3*a^2 - 2*b^2)*E
llipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(5/2))/(10*d*e*(e*Sin[c + d*x])^(5/2))

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Maple [A]  time = 1.971, size = 327, normalized size = 2. \begin{align*}{\frac{1}{d} \left ( -{\frac{4\,ab}{5\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{1}{5\,{e}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) } \left ( \left ( 6\,{a}^{2}-4\,{b}^{2} \right ) \sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+ \left ( -8\,{a}^{2}+2\,{b}^{2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +6\, \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }{a}^{2}-4\, \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }{b}^{2}-3\, \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }{a}^{2}+2\, \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ) \sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \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))^2/(e*sin(d*x+c))^(7/2),x)

[Out]

(-4/5*a*b/e/(e*sin(d*x+c))^(5/2)+1/5/e^3*((6*a^2-4*b^2)*sin(d*x+c)*cos(d*x+c)^4+(-8*a^2+2*b^2)*cos(d*x+c)^2*si
n(d*x+c)+6*sin(d*x+c)^(7/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^
(1/2)*a^2-4*sin(d*x+c)^(7/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))
^(1/2)*b^2-3*sin(d*x+c)^(7/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c)
)^(1/2)*a^2+2*sin(d*x+c)^(7/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c
))^(1/2)*b^2)/sin(d*x+c)^3/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 )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt{e \sin \left (d x + c\right )}}{e^{4} \cos \left (d x + c\right )^{4} - 2 \, e^{4} \cos \left (d x + c\right )^{2} + e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(e*sin(d*x + c))/(e^4*cos(d*x + c)^4 - 2*e^4*cos(
d*x + c)^2 + e^4), 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))**2/(e*sin(d*x+c))**(7/2),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 )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2/(e*sin(d*x + c))^(7/2), x)