Integrand size = 21, antiderivative size = 105 \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {5 d \sin (a+b x)}{6 b \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}+\frac {5 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{12 b} \]
-5/6*d*sin(b*x+a)/b/(d*tan(b*x+a))^(1/2)-1/3*d*sin(b*x+a)^3/b/(d*tan(b*x+a ))^(1/2)-5/12*csc(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Ell ipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^(1/2)*(d*tan(b*x+a))^(1/2 )/b
Result contains complex when optimal does not.
Time = 11.79 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.32 \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {\cos (2 (a+b x)) \sec (a+b x) \left (-5 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sec ^2(a+b x)+(-6+\cos (2 (a+b x))) \sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)}\right ) \sqrt {d \tan (a+b x)}}{6 b \sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \]
-1/6*(Cos[2*(a + b*x)]*Sec[a + b*x]*(-5*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1 )^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sec[a + b*x]^2 + (-6 + Cos[2*(a + b*x)])* Sqrt[Sec[a + b*x]^2]*Sqrt[Tan[a + b*x]])*Sqrt[d*Tan[a + b*x]])/(b*Sqrt[Sec [a + b*x]^2]*Sqrt[Tan[a + b*x]]*(-1 + Tan[a + b*x]^2))
Time = 0.59 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3078, 3042, 3078, 3042, 3081, 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.
\(\displaystyle \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x)^3 \sqrt {d \tan (a+b x)}dx\) |
\(\Big \downarrow \) 3078 |
\(\displaystyle \frac {5}{6} \int \sin (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \int \sin (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3078 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \int \csc (a+b x) \sqrt {d \tan (a+b x)}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \int \frac {\sqrt {d \tan (a+b x)}}{\sin (a+b x)}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {5}{6} \left (\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\sin (a+b x)}}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx}{2 \sqrt {\sin (a+b x)}}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{6} \left (\frac {1}{2} \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \sqrt {d \tan (a+b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{2 b}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\) |
-1/3*(d*Sin[a + b*x]^3)/(b*Sqrt[d*Tan[a + b*x]]) + (5*(-((d*Sin[a + b*x])/ (b*Sqrt[d*Tan[a + b*x]])) + (Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqr t[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(2*b)))/6
3.1.59.3.1 Defintions of rubi rules used
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[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( f*m)), x] + Simp[a^2*((m + n - 1)/m) Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 ] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
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 = 9.25 (sec) , antiderivative size = 1740, normalized size of antiderivative = 16.57
-1/48/b*csc(b*x+a)*(-6*I*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot( b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot (b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+6*I*(cot(b*x+a)-csc(b*x+a))^(1/2)*(- csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi( (1+csc(b*x+a)-cot(b*x+a))^(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))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-c ot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2* 2^(1/2))+6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)* (1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(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))^(1/2)*(-csc (b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+ csc(b*x+a)-cot(b*x+a))^(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))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b* x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/ 2))*cos(b*x+a)-32*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a)) ^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a)) ^(1/2),1/2*2^(1/2))*cos(b*x+a)-8*2^(1/2)*cos(b*x+a)^3*sin(b*x+a)+6*(cot(b* x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot( b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^( 1/2))+6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*...
\[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int { \sqrt {d \tan \left (b x + a\right )} \sin \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\text {Timed out} \]
\[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int { \sqrt {d \tan \left (b x + a\right )} \sin \left (b x + a\right )^{3} \,d x } \]
Exception generated. \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ]ext_reduce Error: Bad Argument TypeDone
Timed out. \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int {\sin \left (a+b\,x\right )}^3\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )} \,d x \]