Integrand size = 21, antiderivative size = 79 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \] Output:
-1/3*d*sin(b*x+a)^3/b/(d*tan(b*x+a))^(3/2)-1/2*EllipticE(cos(a+1/4*Pi+b*x) ,2^(1/2))*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.64 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {\sqrt {d \tan (a+b x)} \left (-\sqrt {\sec ^2(a+b x)} (\sin (a+b x)+\sin (3 (a+b x)))+4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sec (a+b x) \tan (a+b x)\right )}{12 b d \sqrt {\sec ^2(a+b x)}} \] Input:
Integrate[Sin[a + b*x]^3/Sqrt[d*Tan[a + b*x]],x]
Output:
(Sqrt[d*Tan[a + b*x]]*(-(Sqrt[Sec[a + b*x]^2]*(Sin[a + b*x] + Sin[3*(a + b *x)])) + 4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]* Tan[a + b*x]))/(12*b*d*Sqrt[Sec[a + b*x]^2])
Time = 0.42 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3078, 3042, 3081, 3042, 3052, 3042, 3119}
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.
\(\displaystyle \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^3}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3078 |
\(\displaystyle \frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\) |
Input:
Int[Sin[a + b*x]^3/Sqrt[d*Tan[a + b*x]],x]
Output:
-1/3*(d*Sin[a + b*x]^3)/(b*(d*Tan[a + b*x])^(3/2)) + (EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(2*b*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(215\) vs. \(2(70)=140\).
Time = 0.90 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.73
method | result | size |
default | \(\frac {\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-6-6 \sec \left (b x +a \right )\right )+\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (3+3 \sec \left (b x +a \right )\right )+4 \cos \left (b x +a \right )^{3}-10 \cos \left (b x +a \right )+6}{12 b \sqrt {d \tan \left (b x +a \right )}}\) | \(216\) |
Input:
int(sin(b*x+a)^3/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/12/b/(d*tan(b*x+a))^(1/2)*((csc(b*x+a)-cot(b*x+a)+1)^(1/2)*(-2*csc(b*x+a )+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^(1/2)*EllipticE((csc(b*x+ a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(-6-6*sec(b*x+a))+(csc(b*x+a)-cot(b*x+ a)+1)^(1/2)*(-2*csc(b*x+a)+2*cot(b*x+a)+2)^(1/2)*(-csc(b*x+a)+cot(b*x+a))^ (1/2)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))*(3+3*sec(b*x+ a))+4*cos(b*x+a)^3-10*cos(b*x+a)+6)
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\sqrt {d \tan \left (b x + a\right )}} \,d x } \] Input:
integrate(sin(b*x+a)^3/(d*tan(b*x+a))^(1/2),x, algorithm="fricas")
Output:
integral(-(cos(b*x + a)^2 - 1)*sqrt(d*tan(b*x + a))*sin(b*x + a)/(d*tan(b* x + a)), x)
Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\text {Timed out} \] Input:
integrate(sin(b*x+a)**3/(d*tan(b*x+a))**(1/2),x)
Output:
Timed out
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\sqrt {d \tan \left (b x + a\right )}} \,d x } \] Input:
integrate(sin(b*x+a)^3/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)^3/sqrt(d*tan(b*x + a)), x)
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{3}}{\sqrt {d \tan \left (b x + a\right )}} \,d x } \] Input:
integrate(sin(b*x+a)^3/(d*tan(b*x+a))^(1/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)^3/sqrt(d*tan(b*x + a)), x)
Timed out. \[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3}{\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}} \,d x \] Input:
int(sin(a + b*x)^3/(d*tan(a + b*x))^(1/2),x)
Output:
int(sin(a + b*x)^3/(d*tan(a + b*x))^(1/2), x)
\[ \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sin \left (b x +a \right )^{3}}{\tan \left (b x +a \right )}d x \right )}{d} \] Input:
int(sin(b*x+a)^3/(d*tan(b*x+a))^(1/2),x)
Output:
(sqrt(d)*int((sqrt(tan(a + b*x))*sin(a + b*x)**3)/tan(a + b*x),x))/d