Integrand size = 25, antiderivative size = 210 \[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{35 d e}+\frac {2 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{35 d e}-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{35 d e}-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e} \] Output:
-6/35*a*b*(31*a^2+34*b^2)*(e*cos(d*x+c))^(1/2)/d/e+2/7*(7*a^4+28*a^2*b^2+4 *b^4)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d/(e*cos(d*x +c))^(1/2)-2/35*b*(29*a^2+10*b^2)*(e*cos(d*x+c))^(1/2)*(a+b*sin(d*x+c))/d/ e-26/35*a*b*(e*cos(d*x+c))^(1/2)*(a+b*sin(d*x+c))^2/d/e-2/7*b*(e*cos(d*x+c ))^(1/2)*(a+b*sin(d*x+c))^3/d/e
Time = 2.58 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\frac {20 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \cos (c+d x) \left (560 a^3+504 a b^2-56 a b^2 \cos (2 (c+d x))+5 b \left (56 a^2+11 b^2\right ) \sin (c+d x)-5 b^3 \sin (3 (c+d x))\right )}{70 d \sqrt {e \cos (c+d x)}} \] Input:
Integrate[(a + b*Sin[c + d*x])^4/Sqrt[e*Cos[c + d*x]],x]
Output:
(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*C os[c + d*x]])
Time = 1.16 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3171, 27, 3042, 3341, 27, 3042, 3341, 27, 3042, 3148, 3042, 3121, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3171 |
\(\displaystyle \frac {2}{7} \int \frac {(a+b \sin (c+d x))^2 \left (7 a^2+13 b \sin (c+d x) a+6 b^2\right )}{2 \sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \frac {(a+b \sin (c+d x))^2 \left (7 a^2+13 b \sin (c+d x) a+6 b^2\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \frac {(a+b \sin (c+d x))^2 \left (7 a^2+13 b \sin (c+d x) a+6 b^2\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3341 |
\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {(a+b \sin (c+d x)) \left (a \left (35 a^2+82 b^2\right )+3 b \left (29 a^2+10 b^2\right ) \sin (c+d x)\right )}{2 \sqrt {e \cos (c+d x)}}dx-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {(a+b \sin (c+d x)) \left (a \left (35 a^2+82 b^2\right )+3 b \left (29 a^2+10 b^2\right ) \sin (c+d x)\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {(a+b \sin (c+d x)) \left (a \left (35 a^2+82 b^2\right )+3 b \left (29 a^2+10 b^2\right ) \sin (c+d x)\right )}{\sqrt {e \cos (c+d x)}}dx-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3341 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 \left (5 \left (7 a^4+28 b^2 a^2+4 b^4\right )+3 a b \left (31 a^2+34 b^2\right ) \sin (c+d x)\right )}{2 \sqrt {e \cos (c+d x)}}dx-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 \left (7 a^4+28 b^2 a^2+4 b^4\right )+3 a b \left (31 a^2+34 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\int \frac {5 \left (7 a^4+28 b^2 a^2+4 b^4\right )+3 a b \left (31 a^2+34 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}}dx-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {e \cos (c+d x)}}-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {5 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {e \cos (c+d x)}}-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (-\frac {6 a b \left (31 a^2+34 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}-\frac {2 b \left (29 a^2+10 b^2\right ) \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}{d e}+\frac {10 \left (7 a^4+28 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}\right )-\frac {26 a b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^2}{5 d e}\right )-\frac {2 b \sqrt {e \cos (c+d x)} (a+b \sin (c+d x))^3}{7 d e}\) |
Input:
Int[(a + b*Sin[c + d*x])^4/Sqrt[e*Cos[c + d*x]],x]
Output:
(-2*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^3)/(7*d*e) + ((-26*a*b*Sqr t[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(5*d*e) + ((-6*a*b*(31*a^2 + 34* b^2)*Sqrt[e*Cos[c + d*x]])/(d*e) + (10*(7*a^4 + 28*a^2*b^2 + 4*b^4)*Sqrt[C os[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x]]) - (2*b*(2 9*a^2 + 10*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x]))/(d*e))/5)/7
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
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] + Simp[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])
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] + Simp[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])
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] + S imp[1/(m + p + 1) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Sim p[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] && GtQ[m, 0] && !LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && S implerQ[c + d*x, a + b*x])
Leaf count of result is larger than twice the leaf count of optimal. \(411\) vs. \(2(189)=378\).
Time = 15.42 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.96
method | result | size |
default | \(-\frac {2 \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} b^{4}-120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b^{4}+224 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a \,b^{3}-280 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} b^{2}-336 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a \,b^{3}+140 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2} b^{2}+20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{4}+35 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{4}+140 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b^{2}+20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{4}-280 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{3} b -112 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a \,b^{3}+140 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} b +112 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) a \,b^{3}\right )}{35 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\) | \(412\) |
parts | \(\frac {2 a^{4} \sqrt {\cos \left (d x +c \right )}\, \operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d \sqrt {e \cos \left (d x +c \right )}}-\frac {8 b^{4} \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{7 \sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}+\frac {8 a \,b^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-e^{2} \sqrt {e \cos \left (d x +c \right )}\right )}{d \,e^{3}}-\frac {8 a^{3} b \sqrt {e \cos \left (d x +c \right )}}{d e}+\frac {8 b^{2} a^{2} \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}\) | \(505\) |
Input:
int((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-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*cos(1/2*d *x+1/2*c)*sin(1/2*d*x+1/2*c)^8*b^4-120*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2* c)^6*b^4+224*sin(1/2*d*x+1/2*c)^7*a*b^3-280*cos(1/2*d*x+1/2*c)*sin(1/2*d*x +1/2*c)^4*a^2*b^2-336*sin(1/2*d*x+1/2*c)^5*a*b^3+140*cos(1/2*d*x+1/2*c)*si n(1/2*d*x+1/2*c)^2*a^2*b^2+20*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^4+ 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(co s(1/2*d*x+1/2*c),2^(1/2))*b^4-280*sin(1/2*d*x+1/2*c)^3*a^3*b-112*sin(1/2*d *x+1/2*c)^3*a*b^3+140*sin(1/2*d*x+1/2*c)*a^3*b+112*sin(1/2*d*x+1/2*c)*a*b^ 3)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {2 \, {\left (5 \, \sqrt {\frac {1}{2}} {\left (7 i \, a^{4} + 28 i \, a^{2} b^{2} + 4 i \, b^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {\frac {1}{2}} {\left (-7 i \, a^{4} - 28 i \, a^{2} b^{2} - 4 i \, b^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (28 \, a b^{3} \cos \left (d x + c\right )^{2} - 140 \, a^{3} b - 140 \, a b^{3} + 5 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - 14 \, a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{35 \, d e} \] Input:
integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-2/35*(5*sqrt(1/2)*(7*I*a^4 + 28*I*a^2*b^2 + 4*I*b^4)*sqrt(e)*weierstrassP Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(1/2)*(-7*I*a^4 - 28 *I*a^2*b^2 - 4*I*b^4)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I* sin(d*x + c)) - (28*a*b^3*cos(d*x + c)^2 - 140*a^3*b - 140*a*b^3 + 5*(b^4* cos(d*x + c)^2 - 14*a^2*b^2 - 3*b^4)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/( d*e)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+b*sin(d*x+c))**4/(e*cos(d*x+c))**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate((b*sin(d*x + c) + a)^4/sqrt(e*cos(d*x + c)), x)
\[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate((b*sin(d*x + c) + a)^4/sqrt(e*cos(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((a + b*sin(c + d*x))^4/(e*cos(c + d*x))^(1/2),x)
Output:
int((a + b*sin(c + d*x))^4/(e*cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+b \sin (c+d x))^4}{\sqrt {e \cos (c+d x)}} \, dx=\frac {\sqrt {e}\, \left (-8 \sqrt {\cos \left (d x +c \right )}\, a^{3} b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{4} d +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}d x \right ) b^{4} d +4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )}d x \right ) a \,b^{3} d +6 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )}d x \right ) a^{2} b^{2} d \right )}{d e} \] Input:
int((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x)
Output:
(sqrt(e)*( - 8*sqrt(cos(c + d*x))*a**3*b + int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a**4*d + int((sqrt(cos(c + d*x))*sin(c + d*x)**4)/cos(c + d*x),x)* b**4*d + 4*int((sqrt(cos(c + d*x))*sin(c + d*x)**3)/cos(c + d*x),x)*a*b**3 *d + 6*int((sqrt(cos(c + d*x))*sin(c + d*x)**2)/cos(c + d*x),x)*a**2*b**2* d))/(d*e)