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

Optimal. Leaf size=188 \[ \frac{10 e^3 \left (11 a^2+2 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{231 d}+\frac{10 e^4 \left (11 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{e \cos (c+d x)}}+\frac{2 e \left (11 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{5/2}}{77 d}-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e} \]

[Out]

(-26*a*b*(e*Cos[c + d*x])^(9/2))/(99*d*e) + (10*(11*a^2 + 2*b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2])/(231*d*Sqrt[e*Cos[c + d*x]]) + (10*(11*a^2 + 2*b^2)*e^3*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (2*(
11*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*d) - (2*b*(e*Cos[c + d*x])^(9/2)*(a + b*Sin[c + d*x
]))/(11*d*e)

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Rubi [A]  time = 0.191299, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2692, 2669, 2635, 2642, 2641} \[ \frac{10 e^3 \left (11 a^2+2 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{231 d}+\frac{10 e^4 \left (11 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{e \cos (c+d x)}}+\frac{2 e \left (11 a^2+2 b^2\right ) \sin (c+d x) (e \cos (c+d x))^{5/2}}{77 d}-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-26*a*b*(e*Cos[c + d*x])^(9/2))/(99*d*e) + (10*(11*a^2 + 2*b^2)*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2])/(231*d*Sqrt[e*Cos[c + d*x]]) + (10*(11*a^2 + 2*b^2)*e^3*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (2*(
11*a^2 + 2*b^2)*e*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(77*d) - (2*b*(e*Cos[c + d*x])^(9/2)*(a + b*Sin[c + d*x
]))/(11*d*e)

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (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 2635

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

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 (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2 \, dx &=-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}+\frac{2}{11} \int (e \cos (c+d x))^{7/2} \left (\frac{11 a^2}{2}+b^2+\frac{13}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}+\frac{1}{11} \left (11 a^2+2 b^2\right ) \int (e \cos (c+d x))^{7/2} \, dx\\ &=-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}+\frac{2 \left (11 a^2+2 b^2\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}+\frac{1}{77} \left (5 \left (11 a^2+2 b^2\right ) e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}+\frac{10 \left (11 a^2+2 b^2\right ) e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac{2 \left (11 a^2+2 b^2\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}+\frac{1}{231} \left (5 \left (11 a^2+2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}+\frac{10 \left (11 a^2+2 b^2\right ) e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac{2 \left (11 a^2+2 b^2\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}+\frac{\left (5 \left (11 a^2+2 b^2\right ) e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{231 \sqrt{e \cos (c+d x)}}\\ &=-\frac{26 a b (e \cos (c+d x))^{9/2}}{99 d e}+\frac{10 \left (11 a^2+2 b^2\right ) e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d \sqrt{e \cos (c+d x)}}+\frac{10 \left (11 a^2+2 b^2\right ) e^3 \sqrt{e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac{2 \left (11 a^2+2 b^2\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{77 d}-\frac{2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{11 d e}\\ \end{align*}

Mathematica [A]  time = 1.90765, size = 160, normalized size = 0.85 \[ \frac{(e \cos (c+d x))^{7/2} \left (40 \left (11 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{1}{6} \sqrt{\cos (c+d x)} \left (6 \left (572 a^2+41 b^2\right ) \sin (c+d x)+8 \cos (2 (c+d x)) \left (9 \left (11 a^2-5 b^2\right ) \sin (c+d x)-154 a b\right )-14 b \cos (4 (c+d x)) (22 a+9 b \sin (c+d x))\right )-154 a b \sqrt{\cos (c+d x)}\right )}{924 d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(7/2)*(-154*a*b*Sqrt[Cos[c + d*x]] + 40*(11*a^2 + 2*b^2)*EllipticF[(c + d*x)/2, 2] + (Sqrt[C
os[c + d*x]]*(6*(572*a^2 + 41*b^2)*Sin[c + d*x] - 14*b*Cos[4*(c + d*x)]*(22*a + 9*b*Sin[c + d*x]) + 8*Cos[2*(c
+ d*x)]*(-154*a*b + 9*(11*a^2 - 5*b^2)*Sin[c + d*x])))/6))/(924*d*Cos[c + d*x]^(7/2))

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Maple [B]  time = 1.214, size = 473, normalized size = 2.5 \begin{align*} -{\frac{2\,{e}^{4}}{693\,d} \left ( -4032\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}-4928\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}+10080\,{b}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) +1584\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+12320\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}-9792\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-2376\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-12320\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+4608\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+1848\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+6160\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-924\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+165\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}+30\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{2}-528\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1540\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+30\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+154\,\sin \left ( 1/2\,dx+c/2 \right ) ab \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/693/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(-4032*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12-4928*a*b*sin(1/2*d*x+1/2*c)^11+10080*b^2*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+1584*a^2*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^8+12320*a*b*sin(1/2*d*x+1/2*c)^9-9792*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2
376*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12320*a*b*sin(1/2*d*x+1/2*c)^7+4608*b^2*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^6+1848*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6160*a*b*sin(1/2*d*x+1/2*c)^5-924*b^2*cos(1
/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+165*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+30*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*b^2-528*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-1540*a*b*sin(1/2*d*x+1/2*c)^3+30
*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+154*sin(1/2*d*x+1/2*c)*a*b)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*e^3*cos(d*x + c)^5 - 2*a*b*e^3*cos(d*x + c)^3*sin(d*x + c) - (a^2 + b^2)*e^3*cos(d*x + c)^3)*sq
rt(e*cos(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((e*cos(d*x+c))**(7/2)*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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