Integrand size = 19, antiderivative size = 79 \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b d \sqrt {d \tan (a+b x)}} \] Output:
sin(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)+1/2*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/ 2))*sec(b*x+a)*sin(2*b*x+2*a)^(1/2)/b/d/(d*tan(b*x+a))^(1/2)
Result contains complex when optimal does not.
Time = 0.62 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.59 \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\cos (2 (a+b x)) \sec (a+b x) \left (\sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sec ^2(a+b x)-\sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)}\right ) \tan ^{\frac {3}{2}}(a+b x)}{b \sqrt {\sec ^2(a+b x)} (d \tan (a+b x))^{3/2} \left (-1+\tan ^2(a+b x)\right )} \] Input:
Integrate[Sin[a + b*x]/(d*Tan[a + b*x])^(3/2),x]
Output:
(Cos[2*(a + b*x)]*Sec[a + b*x]*((-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)* Sqrt[Tan[a + b*x]]], -1]*Sec[a + b*x]^2 - Sqrt[Sec[a + b*x]^2]*Sqrt[Tan[a + b*x]])*Tan[a + b*x]^(3/2))/(b*Sqrt[Sec[a + b*x]^2]*(d*Tan[a + b*x])^(3/2 )*(-1 + Tan[a + b*x]^2))
Time = 0.46 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3082, 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.
\(\displaystyle \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3082 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sqrt {\sin (a+b x)}}dx}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \int \frac {\cos (a+b x)^{3/2}}{\sqrt {\sin (a+b x)}}dx}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3049 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \left (\frac {1}{2} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx+\frac {\sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}{b}\right )}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \left (\frac {1}{2} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}}dx+\frac {\sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}{b}\right )}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \left (\frac {\sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}{b}\right )}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \left (\frac {\sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}{b}\right )}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sqrt {\sin (a+b x)} \left (\frac {\sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}{b}+\frac {\sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {\sin (a+b x)} \sqrt {\cos (a+b x)}}\right )}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
Input:
Int[Sin[a + b*x]/(d*Tan[a + b*x])^(3/2),x]
Output:
(Sqrt[Sin[a + b*x]]*((Sqrt[Cos[a + b*x]]*Sqrt[Sin[a + b*x]])/b + (Elliptic F[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(2*b*Sqrt[Cos[a + b*x]]*Sqrt[ Sin[a + b*x]])))/(d*Sqrt[Cos[a + b*x]]*Sqrt[d*Tan[a + b*x]])
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[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* (a*Sin[e + f*x])^(n + 1))) 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]
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]
Time = 2.74 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.44
method | result | size |
default | \(\frac {\frac {\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (b x +a \right )\right )}{2}+\sin \left (b x +a \right )}{b \sqrt {d \tan \left (b x +a \right )}\, d}\) | \(114\) |
Input:
int(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/b*(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)*EllipticF((csc(b*x+a)-cot(b*x+a)+1)^(1/ 2),1/2*2^(1/2))*(1+sec(b*x+a))+sin(b*x+a))/(d*tan(b*x+a))^(1/2)/d
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
Output:
integral(sqrt(d*tan(b*x + a))*sin(b*x + a)/(d^2*tan(b*x + a)^2), x)
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\sin {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sin(b*x+a)/(d*tan(b*x+a))**(3/2),x)
Output:
Integral(sin(a + b*x)/(d*tan(a + b*x))**(3/2), x)
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)/(d*tan(b*x + a))^(3/2), x)
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int { \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)/(d*tan(b*x + a))^(3/2), x)
Timed out. \[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\sin \left (a+b\,x\right )}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \] Input:
int(sin(a + b*x)/(d*tan(a + b*x))^(3/2),x)
Output:
int(sin(a + b*x)/(d*tan(a + b*x))^(3/2), x)
\[ \int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sin \left (b x +a \right )}{\tan \left (b x +a \right )^{2}}d x \right )}{d^{2}} \] Input:
int(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x)
Output:
(sqrt(d)*int((sqrt(tan(a + b*x))*sin(a + b*x))/tan(a + b*x)**2,x))/d**2