3.568 \(\int \frac{(a+b \sin (c+d x))^4}{\sqrt{e \cos (c+d x)}} \, dx\)

Optimal. Leaf size=210 \[ -\frac{6 a b \left (31 a^2+34 b^2\right ) \sqrt{e \cos (c+d x)}}{35 d e}-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}+\frac{2 \left (28 a^2 b^2+7 a^4+4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d \sqrt{e \cos (c+d x)}}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e} \]

[Out]

(-6*a*b*(31*a^2 + 34*b^2)*Sqrt[e*Cos[c + d*x]])/(35*d*e) + (2*(7*a^4 + 28*a^2*b^2 + 4*b^4)*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2])/(7*d*Sqrt[e*Cos[c + d*x]]) - (2*b*(29*a^2 + 10*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*Sin
[c + d*x]))/(35*d*e) - (26*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(35*d*e) - (2*b*Sqrt[e*Cos[c + d*x
]]*(a + b*Sin[c + d*x])^3)/(7*d*e)

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Rubi [A]  time = 0.445017, antiderivative size = 210, 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, 2862, 2669, 2642, 2641} \[ -\frac{6 a b \left (31 a^2+34 b^2\right ) \sqrt{e \cos (c+d x)}}{35 d e}-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}+\frac{2 \left (28 a^2 b^2+7 a^4+4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d \sqrt{e \cos (c+d x)}}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^4/Sqrt[e*Cos[c + d*x]],x]

[Out]

(-6*a*b*(31*a^2 + 34*b^2)*Sqrt[e*Cos[c + d*x]])/(35*d*e) + (2*(7*a^4 + 28*a^2*b^2 + 4*b^4)*Sqrt[Cos[c + d*x]]*
EllipticF[(c + d*x)/2, 2])/(7*d*Sqrt[e*Cos[c + d*x]]) - (2*b*(29*a^2 + 10*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*Sin
[c + d*x]))/(35*d*e) - (26*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(35*d*e) - (2*b*Sqrt[e*Cos[c + d*x
]]*(a + b*Sin[c + d*x])^3)/(7*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 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 \sin (c+d x))^4}{\sqrt{e \cos (c+d x)}} \, dx &=-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}+\frac{2}{7} \int \frac{(a+b \sin (c+d x))^2 \left (\frac{7 a^2}{2}+3 b^2+\frac{13}{2} a b \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}+\frac{4}{35} \int \frac{(a+b \sin (c+d x)) \left (\frac{1}{4} a \left (35 a^2+82 b^2\right )+\frac{3}{4} b \left (29 a^2+10 b^2\right ) \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}+\frac{8}{105} \int \frac{\frac{15}{8} \left (7 a^4+28 a^2 b^2+4 b^4\right )+\frac{9}{8} a b \left (31 a^2+34 b^2\right ) \sin (c+d x)}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{6 a b \left (31 a^2+34 b^2\right ) \sqrt{e \cos (c+d x)}}{35 d e}-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}+\frac{1}{7} \left (7 a^4+28 a^2 b^2+4 b^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{6 a b \left (31 a^2+34 b^2\right ) \sqrt{e \cos (c+d x)}}{35 d e}-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}+\frac{\left (\left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 \sqrt{e \cos (c+d x)}}\\ &=-\frac{6 a b \left (31 a^2+34 b^2\right ) \sqrt{e \cos (c+d x)}}{35 d e}+\frac{2 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d \sqrt{e \cos (c+d x)}}-\frac{2 b \left (29 a^2+10 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}-\frac{26 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac{2 b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\\ \end{align*}

Mathematica [A]  time = 1.09054, size = 130, normalized size = 0.62 \[ \frac{20 \left (28 a^2 b^2+7 a^4+4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-b \cos (c+d x) \left (5 b \left (56 a^2+11 b^2\right ) \sin (c+d x)+560 a^3-56 a b^2 \cos (2 (c+d x))+504 a b^2-5 b^3 \sin (3 (c+d x))\right )}{70 d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^4/Sqrt[e*Cos[c + d*x]],x]

[Out]

(20*(7*a^4 + 28*a^2*b^2 + 4*b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - b*Cos[c + d*x]*(560*a^3 + 504*
a*b^2 - 56*a*b^2*Cos[2*(c + d*x)] + 5*b*(56*a^2 + 11*b^2)*Sin[c + d*x] - 5*b^3*Sin[3*(c + d*x)]))/(70*d*Sqrt[e
*Cos[c + d*x]])

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Maple [A]  time = 1.437, size = 412, normalized size = 2. \begin{align*} -{\frac{2}{35\,d} \left ( 80\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+224\,a{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-120\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-280\,{a}^{2}{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-336\,a{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+35\,\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}^{4}+140\,\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}{b}^{2}+20\,\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}^{4}-280\,{a}^{3}b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+140\,{a}^{2}{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-112\,a{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+20\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+140\,{a}^{3}b\sin \left ( 1/2\,dx+c/2 \right ) +112\,a{b}^{3}\sin \left ( 1/2\,dx+c/2 \right ) \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((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x)

[Out]

-2/35/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(80*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+2
24*a*b^3*sin(1/2*d*x+1/2*c)^7-120*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-280*a^2*b^2*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^4-336*a*b^3*sin(1/2*d*x+1/2*c)^5+35*(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^4+140*(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*b^2+20*(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))*b^4-280*a^3*b*sin(1/2*d*x+1/2*c)^3+140*a^2*b^2*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^2-112*a*b^3*sin(1/2*d*x+1/2*c)^3+20*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+140*a^3*b*si
n(1/2*d*x+1/2*c)+112*a*b^3*sin(1/2*d*x+1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^4/sqrt(e*cos(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}}{e \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^4/sqrt(e*cos(d*x + c)), x)