Integrand size = 25, antiderivative size = 315 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {c^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} b d^{5/2}}-\frac {c^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} b d^{5/2}}-\frac {c^{5/2} \log \left (\sqrt {c}-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{2 \sqrt {2} b d^{5/2}}+\frac {c^{5/2} \log \left (\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+\sqrt {c} \tan (a+b x)\right )}{2 \sqrt {2} b d^{5/2}}+\frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}} \]
2/3*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(3/2)+1/2*c^(5/2)*arctan(1-2 ^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/c^(1/2)/(d*cos(b*x+a))^(1/2))/b/d^(5/2 )*2^(1/2)-1/2*c^(5/2)*arctan(1+2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/c^(1/2 )/(d*cos(b*x+a))^(1/2))/b/d^(5/2)*2^(1/2)-1/4*c^(5/2)*ln(c^(1/2)-2^(1/2)*d ^(1/2)*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)+c^(1/2)*tan(b*x+a))/b/d^( 5/2)*2^(1/2)+1/4*c^(5/2)*ln(c^(1/2)+2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/( d*cos(b*x+a))^(1/2)+c^(1/2)*tan(b*x+a))/b/d^(5/2)*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.21 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {7}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{7/2}}{7 b c d (d \cos (a+b x))^{3/2}} \]
(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[7/4, 7/4, 11/4, Sin[a + b*x]^2 ]*(c*Sin[a + b*x])^(7/2))/(7*b*c*d*(d*Cos[a + b*x])^(3/2))
Time = 0.58 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3046, 3042, 3054, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \int \frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}dx}{d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \int \frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}dx}{d^2}\) |
\(\Big \downarrow \) 3054 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \int \frac {c \tan (a+b x)}{d \left (\tan ^2(a+b x) c^2+c^2\right )}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{b d}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\int \frac {\tan (a+b x) c+c}{\tan ^2(a+b x) c^2+c^2}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}-\frac {\int \frac {c-c \tan (a+b x)}{\tan ^2(a+b x) c^2+c^2}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\int \frac {1}{\frac {\tan (a+b x) c}{d}+\frac {c}{d}-\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}+\frac {\int \frac {1}{\frac {\tan (a+b x) c}{d}+\frac {c}{d}+\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}}{2 d}-\frac {\int \frac {c-c \tan (a+b x)}{\tan ^2(a+b x) c^2+c^2}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\int \frac {1}{-\frac {c \tan (a+b x)}{d}-1}d\left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\int \frac {1}{-\frac {c \tan (a+b x)}{d}-1}d\left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\int \frac {c-c \tan (a+b x)}{\tan ^2(a+b x) c^2+c^2}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\int \frac {c-c \tan (a+b x)}{\tan ^2(a+b x) c^2+c^2}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {c}-\frac {2 \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{\sqrt {d} \left (\frac {\tan (a+b x) c}{d}+\frac {c}{d}-\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}\right )}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{\sqrt {d} \left (\frac {\tan (a+b x) c}{d}+\frac {c}{d}+\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}\right )}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {c}-\frac {2 \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{\sqrt {d} \left (\frac {\tan (a+b x) c}{d}+\frac {c}{d}-\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}\right )}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {2} \sqrt {c} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}\right )}{\sqrt {d} \left (\frac {\tan (a+b x) c}{d}+\frac {c}{d}+\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}\right )}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {c}-\frac {2 \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{\frac {\tan (a+b x) c}{d}+\frac {c}{d}-\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {2} \sqrt {c} d}+\frac {\int \frac {\sqrt {c}+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{\frac {\tan (a+b x) c}{d}+\frac {c}{d}+\frac {\sqrt {2} \sqrt {c \sin (a+b x)} \sqrt {c}}{\sqrt {d} \sqrt {d \cos (a+b x)}}}d\frac {\sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}}{2 \sqrt {c} d}}{2 d}\right )}{b d}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 c^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\frac {\log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+c \tan (a+b x)+c\right )}{2 \sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+c \tan (a+b x)+c\right )}{2 \sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}\right )}{b d}\) |
(-2*c^3*((-(ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sqr t[d*Cos[a + b*x]])]/(Sqrt[2]*Sqrt[c]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[ d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sqrt[d*Cos[a + b*x]])]/(Sqrt[2]*Sqrt[c]* Sqrt[d]))/(2*d) - (-1/2*Log[c - (Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[c*Sin[a + b* x]])/Sqrt[d*Cos[a + b*x]] + c*Tan[a + b*x]]/(Sqrt[2]*Sqrt[c]*Sqrt[d]) + Lo g[c + (Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/Sqrt[d*Cos[a + b*x]] + c*Tan[a + b*x]]/(2*Sqrt[2]*Sqrt[c]*Sqrt[d]))/(2*d)))/(b*d) + (2*c*(c*Sin [a + b*x])^(3/2))/(3*b*d*(d*Cos[a + b*x])^(3/2))
3.3.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Sin[e + f *x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> With[{k = Denominator[m]}, Simp[k*a*(b/f) Subst[Int[x^(k *(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]
Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(238)=476\).
Time = 0.26 (sec) , antiderivative size = 492, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {\sqrt {2}\, \left (6 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right ) \cos \left (b x +a \right )+6 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}-\cos \left (b x +a \right )+1}{\cos \left (b x +a \right )-1}\right ) \cos \left (b x +a \right )-3 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \ln \left (2 \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \cot \left (b x +a \right )+2 \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \csc \left (b x +a \right )+2-2 \cot \left (b x +a \right )\right ) \cos \left (b x +a \right )+3 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \ln \left (-2 \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \cot \left (b x +a \right )-2 \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (1+\cos \left (b x +a \right )\right )^{2}}}\, \csc \left (b x +a \right )+2-2 \cot \left (b x +a \right )\right ) \cos \left (b x +a \right )-4 \sqrt {2}\, \cos \left (b x +a \right )+4 \sqrt {2}\right ) c^{2} \sqrt {c \sin \left (b x +a \right )}\, \left (1+\cos \left (b x +a \right )\right ) \sec \left (b x +a \right ) \csc \left (b x +a \right )}{12 b \sqrt {d \cos \left (b x +a \right )}\, d^{2}}\) | \(492\) |
1/12/b*2^(1/2)*(6*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a))^2)^(1/2)*arctan(( sin(b*x+a)*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a))^2)^(1/2)+cos(b*x +a)-1)/(cos(b*x+a)-1))*cos(b*x+a)+6*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a)) ^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a) )^2)^(1/2)-cos(b*x+a)+1)/(cos(b*x+a)-1))*cos(b*x+a)-3*(-sin(b*x+a)*cos(b*x +a)/(1+cos(b*x+a))^2)^(1/2)*ln(2*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b* x+a))^2)^(1/2)*cot(b*x+a)+2*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a)) ^2)^(1/2)*csc(b*x+a)+2-2*cot(b*x+a))*cos(b*x+a)+3*(-sin(b*x+a)*cos(b*x+a)/ (1+cos(b*x+a))^2)^(1/2)*ln(-2*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a ))^2)^(1/2)*cot(b*x+a)-2*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(1+cos(b*x+a))^2) ^(1/2)*csc(b*x+a)+2-2*cot(b*x+a))*cos(b*x+a)-4*2^(1/2)*cos(b*x+a)+4*2^(1/2 ))*c^2*(c*sin(b*x+a))^(1/2)*(1+cos(b*x+a))/(d*cos(b*x+a))^(1/2)/d^2*sec(b* x+a)*csc(b*x+a)
Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 1164, normalized size of antiderivative = 3.70 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\text {Too large to display} \]
1/24*(3*b*d^3*(-c^10/(b^4*d^10))^(1/4)*cos(b*x + a)^2*log(-1/2*c^8*cos(b*x + a)*sin(b*x + a) + 1/2*(b^3*d^7*(-c^10/(b^4*d^10))^(3/4)*cos(b*x + a) - b*c^5*d^2*(-c^10/(b^4*d^10))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt (c*sin(b*x + a)) + 1/4*(2*b^2*c^3*d^5*cos(b*x + a)^2 - b^2*c^3*d^5)*sqrt(- c^10/(b^4*d^10))) - 3*b*d^3*(-c^10/(b^4*d^10))^(1/4)*cos(b*x + a)^2*log(-1 /2*c^8*cos(b*x + a)*sin(b*x + a) - 1/2*(b^3*d^7*(-c^10/(b^4*d^10))^(3/4)*c os(b*x + a) - b*c^5*d^2*(-c^10/(b^4*d^10))^(1/4)*sin(b*x + a))*sqrt(d*cos( b*x + a))*sqrt(c*sin(b*x + a)) + 1/4*(2*b^2*c^3*d^5*cos(b*x + a)^2 - b^2*c ^3*d^5)*sqrt(-c^10/(b^4*d^10))) - 3*I*b*d^3*(-c^10/(b^4*d^10))^(1/4)*cos(b *x + a)^2*log(-1/2*c^8*cos(b*x + a)*sin(b*x + a) + 1/2*(I*b^3*d^7*(-c^10/( b^4*d^10))^(3/4)*cos(b*x + a) + I*b*c^5*d^2*(-c^10/(b^4*d^10))^(1/4)*sin(b *x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - 1/4*(2*b^2*c^3*d^5*co s(b*x + a)^2 - b^2*c^3*d^5)*sqrt(-c^10/(b^4*d^10))) + 3*I*b*d^3*(-c^10/(b^ 4*d^10))^(1/4)*cos(b*x + a)^2*log(-1/2*c^8*cos(b*x + a)*sin(b*x + a) + 1/2 *(-I*b^3*d^7*(-c^10/(b^4*d^10))^(3/4)*cos(b*x + a) - I*b*c^5*d^2*(-c^10/(b ^4*d^10))^(1/4)*sin(b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a)) - 1/4*(2*b^2*c^3*d^5*cos(b*x + a)^2 - b^2*c^3*d^5)*sqrt(-c^10/(b^4*d^10))) - 3*b*d^3*(-c^10/(b^4*d^10))^(1/4)*cos(b*x + a)^2*log(c^8 + 2*(b^3*d^7*(-c^ 10/(b^4*d^10))^(3/4)*sin(b*x + a) - b*c^5*d^2*(-c^10/(b^4*d^10))^(1/4)*cos (b*x + a))*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))) + 3*b*d^3*(-c^10/...
Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \]