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

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

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.47, 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]

-1/10*(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)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(5/2))/(d*e*(e*Sin[c + d*x])^(5/2))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {e \sin \left (d x + c\right )}}{e^{4} \cos \left (d x + c\right )^{4} - 2 \, e^{4} \cos \left (d x + c\right )^{2} + e^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.26, size = 327, normalized size = 1.98 \[ \frac {-\frac {4 a b}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\left (6 a^{2}-4 b^{2}\right ) \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+\left (-8 a^{2}+2 b^{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

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)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2
^(1/2))*a^2-4*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/
2*2^(1/2))*b^2-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2)
,1/2*2^(1/2))*a^2+2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1
/2),1/2*2^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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