Integrand size = 19, antiderivative size = 76 \[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {3 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b} \] Output:
3*d^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)+2*d*sin(b*x+a)*(d*tan(b*x+a))^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76 \[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {2 \cos (a+b x) \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) (d \tan (a+b x))^{3/2}}{b} \] Input:
Integrate[Sin[a + b*x]*(d*Tan[a + b*x])^(3/2),x]
Output:
(-2*Cos[a + b*x]*(-1 + Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*S qrt[Sec[a + b*x]^2])*(d*Tan[a + b*x])^(3/2))/b
Time = 0.41 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3074, 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.
\(\displaystyle \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (a+b x) (d \tan (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3074 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-3 d^2 \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-3 d^2 \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {3 d^2 \sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {3 d^2 \sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3052 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {3 d^2 \sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {3 d^2 \sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 d \sin (a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {3 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\) |
Input:
Int[Sin[a + b*x]*(d*Tan[a + b*x])^(3/2),x]
Output:
(-3*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*Sqrt[Sin[2*a + 2*b*x ]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Sin[a + b*x]*Sqrt[d*Tan[a + b*x]])/b
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*(n - 1))), x] - Simp[b^2*((m + n - 1)/(n - 1)) Int[(a*Sin[e + f*x])^m*(b*Ta n[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && In tegersQ[2*m, 2*n] && !(GtQ[m, 1] && !IntegerQ[(m - 1)/2])
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. \(277\) vs. \(2(71)=142\).
Time = 1.05 (sec) , antiderivative size = 278, normalized size of antiderivative = 3.66
method | result | size |
default | \(\frac {\csc \left (b x +a \right ) \left (\left (6 \cos \left (b x +a \right )+6\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-3 \cos \left (b x +a \right )-3\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \cos \left (b x +a \right )^{2}-6 \cos \left (b x +a \right )+4\right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {d \tan \left (b x +a \right )}\, d \sqrt {2}}{4 b \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\) | \(278\) |
Input:
int(sin(b*x+a)*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4/b*csc(b*x+a)*((6*cos(b*x+a)+6)*(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((cs c(b*x+a)-cot(b*x+a)+1)^(1/2),1/2*2^(1/2))+(-3*cos(b*x+a)-3)*(csc(b*x+a)-co t(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))+2*cos(b *x+a)^2-6*cos(b*x+a)+4)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)* (d*tan(b*x+a))^(1/2)*d/(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*2^( 1/2)
\[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
Output:
integral(sqrt(d*tan(b*x + a))*d*sin(b*x + a)*tan(b*x + a), x)
\[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sin {\left (a + b x \right )}\, dx \] Input:
integrate(sin(b*x+a)*(d*tan(b*x+a))**(3/2),x)
Output:
Integral((d*tan(a + b*x))**(3/2)*sin(a + b*x), x)
\[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(sin(b*x+a)*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
Output:
integrate((d*tan(b*x + a))^(3/2)*sin(b*x + a), x)
Exception generated. \[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sin(b*x+a)*(d*tan(b*x+a))^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo r the root of a polynomial with parameters. This might be wrong.The choice was done
Timed out. \[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \sin \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \] 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)
\[ \int \sin (a+b x) (d \tan (a+b x))^{3/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sin \left (b x +a \right ) \tan \left (b x +a \right )d x \right ) d \] 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),x)*d