Integrand size = 25, antiderivative size = 131 \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\frac {c d \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}{6 b}-\frac {c (d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}}{3 b d}+\frac {c^2 d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{12 b \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \]
-1/3*c*(d*cos(b*x+a))^(5/2)*(c*sin(b*x+a))^(1/2)/b/d+1/6*c*d*(d*cos(b*x+a) )^(1/2)*(c*sin(b*x+a))^(1/2)/b-1/12*c^2*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/si n(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^(1/2)/ b/(d*cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.54 \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\frac {2 c d \sqrt {d \cos (a+b x)} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {9}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan ^2(a+b x)}{5 b} \]
(2*c*d*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 5/4, 9/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]]*Tan[a + b*x]^2)/(5*b)
Time = 0.58 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3048, 3042, 3049, 3042, 3053, 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.
\(\displaystyle \int (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {1}{6} c^2 \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}}dx-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} c^2 \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}}dx-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3049 |
\(\displaystyle \frac {1}{6} c^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} c^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\) |
-1/3*(c*(d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]])/(b*d) + (c^2*((d*Sqrt [d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])/(b*c) + (d^2*EllipticF[a - Pi/4 + b *x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(2*b*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b* x]])))/6
3.3.67.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(b*Sin[e + f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/ (b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Sin[e + f*x])^n*(a *Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, 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]
Result contains complex when optimal does not.
Time = 1.47 (sec) , antiderivative size = 1744, normalized size of antiderivative = 13.31
-1/48/b*2^(1/2)*(6*I*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+ a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+csc(b*x+ a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(b*x+a)-6*I*(-cot(b*x+a)+csc(b*x+a)+ 1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*Ell ipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))+8*2^(1/2)* cos(b*x+a)^3*sin(b*x+a)-6*I*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-c sc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+c sc(b*x+a)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a)+6*I*(-cot(b*x+a)+csc( b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1 /2)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))-6*( -cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a )-csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2-1/2*I, 1/2*2^(1/2))*cos(b*x+a)+8*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc (b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+csc( b*x+a)+1)^(1/2),1/2*2^(1/2))*cos(b*x+a)-6*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2) *(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi( (-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(b*x+a)-6*(-cot (b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-cs c(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2-1/2*I,1/2* 2^(1/2))+8*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(...
\[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
\[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} \left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\int {\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2} \,d x \]