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

Optimal. Leaf size=178 \[ -\frac {78 a^4 \sqrt {e \cos (c+d x)}}{7 d e}-\frac {78 \left (a^4 \sin (c+d x)+a^4\right ) \sqrt {e \cos (c+d x)}}{35 d e}+\frac {78 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {26 \left (a^2 \sin (c+d x)+a^2\right )^2 \sqrt {e \cos (c+d x)}}{35 d e}-\frac {2 a (a \sin (c+d x)+a)^3 \sqrt {e \cos (c+d x)}}{7 d e} \]

[Out]

78/7*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2
)/d/(e*cos(d*x+c))^(1/2)-78/7*a^4*(e*cos(d*x+c))^(1/2)/d/e-2/7*a*(a+a*sin(d*x+c))^3*(e*cos(d*x+c))^(1/2)/d/e-2
6/35*(a^2+a^2*sin(d*x+c))^2*(e*cos(d*x+c))^(1/2)/d/e-78/35*(a^4+a^4*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/d/e

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Rubi [A]  time = 0.21, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2678, 2669, 2642, 2641} \[ -\frac {78 a^4 \sqrt {e \cos (c+d x)}}{7 d e}-\frac {26 \left (a^2 \sin (c+d x)+a^2\right )^2 \sqrt {e \cos (c+d x)}}{35 d e}-\frac {78 \left (a^4 \sin (c+d x)+a^4\right ) \sqrt {e \cos (c+d x)}}{35 d e}+\frac {78 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {2 a (a \sin (c+d x)+a)^3 \sqrt {e \cos (c+d x)}}{7 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-78*a^4*Sqrt[e*Cos[c + d*x]])/(7*d*e) + (78*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*d*Sqrt[e*Cos
[c + d*x]]) - (2*a*Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^3)/(7*d*e) - (26*Sqrt[e*Cos[c + d*x]]*(a^2 + a^2*
Sin[c + d*x])^2)/(35*d*e) - (78*Sqrt[e*Cos[c + d*x]]*(a^4 + a^4*Sin[c + d*x]))/(35*d*e)

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]

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 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 2678

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[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx &=-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}+\frac {1}{7} (13 a) \int \frac {(a+a \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}-\frac {26 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )^2}{35 d e}+\frac {1}{35} \left (117 a^2\right ) \int \frac {(a+a \sin (c+d x))^2}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}-\frac {26 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )^2}{35 d e}-\frac {78 \sqrt {e \cos (c+d x)} \left (a^4+a^4 \sin (c+d x)\right )}{35 d e}+\frac {1}{7} \left (39 a^3\right ) \int \frac {a+a \sin (c+d x)}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {78 a^4 \sqrt {e \cos (c+d x)}}{7 d e}-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}-\frac {26 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )^2}{35 d e}-\frac {78 \sqrt {e \cos (c+d x)} \left (a^4+a^4 \sin (c+d x)\right )}{35 d e}+\frac {1}{7} \left (39 a^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {78 a^4 \sqrt {e \cos (c+d x)}}{7 d e}-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}-\frac {26 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )^2}{35 d e}-\frac {78 \sqrt {e \cos (c+d x)} \left (a^4+a^4 \sin (c+d x)\right )}{35 d e}+\frac {\left (39 a^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 \sqrt {e \cos (c+d x)}}\\ &=-\frac {78 a^4 \sqrt {e \cos (c+d x)}}{7 d e}+\frac {78 a^4 \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 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3}{7 d e}-\frac {26 \sqrt {e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )^2}{35 d e}-\frac {78 \sqrt {e \cos (c+d x)} \left (a^4+a^4 \sin (c+d x)\right )}{35 d e}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 64, normalized size = 0.36 \[ -\frac {32 \sqrt [4]{2} a^4 \sqrt {e \cos (c+d x)} \, _2F_1\left (-\frac {13}{4},\frac {1}{4};\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt [4]{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*2^(1/4)*a^4*Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[-13/4, 1/4, 5/4, (1 - Sin[c + d*x])/2])/(d*e*(1 + Sin[
c + d*x])^(1/4))

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} \cos \left (d x + c\right )^{4} - 8 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} - 4 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c))*sqr
t(e*cos(d*x + c))/(e*cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.86, size = 222, normalized size = 1.25 \[ -\frac {2 a^{4} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+224 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-280 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-336 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+160 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-392 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+252 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*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)*a^4*(80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-1
20*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+224*sin(1/2*d*x+1/2*c)^7-280*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c
)-336*sin(1/2*d*x+1/2*c)^5+195*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)+160*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-392*sin(1/2*d*x+1/2*c)^3+252*sin(1/2*d*x+1/2*
c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(1/2),x)

[Out]

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(1/2), 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+a*sin(d*x+c))**4/(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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