\(\int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [69]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac {7 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b} \] Output:

7/3*d^3*sin(b*x+a)^3/b/(d*tan(b*x+a))^(3/2)+7/2*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)^3*(d*tan(b*x+a))^(1/2)/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.82 \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {\left (-28 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\tan ^2(a+b x)\right ) \sec (a+b x)+2 \cos (a+b x) (13+\cos (2 (a+b x))) \sqrt {\sec ^2(a+b x)}\right ) (d \tan (a+b x))^{3/2}}{12 b \sqrt {\sec ^2(a+b x)}} \] Input:

Integrate[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]
 

Output:

((-28*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x] + 2*C 
os[a + b*x]*(13 + Cos[2*(a + b*x)])*Sqrt[Sec[a + b*x]^2])*(d*Tan[a + b*x]) 
^(3/2))/(12*b*Sqrt[Sec[a + b*x]^2])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3074, 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.

\(\displaystyle \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (a+b x)^3 (d \tan (a+b x))^{3/2}dx\)

\(\Big \downarrow \) 3074

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \int \frac {\sin (a+b x)^3}{\sqrt {d \tan (a+b x)}}dx\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}}dx-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3081

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {\sin (a+b x) \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-7 d^2 \left (\frac {\sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}\right )\)

Input:

Int[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]
 

Output:

(2*d*Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/b - 7*d^2*(-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]]))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3052

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]
 

rule 3074

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])
 

rule 3078

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]
 

rule 3081

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])
 

rule 3119

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(99)=198\).

Time = 1.40 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {\csc \left (b x +a \right ) \left (\left (21 \cos \left (b x +a \right )+21\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-42 \cos \left (b x +a \right )-42\right ) \sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}\, \sqrt {-2 \csc \left (b x +a \right )+2 \cot \left (b x +a \right )+2}\, \sqrt {-\csc \left (b x +a \right )+\cot \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\csc \left (b x +a \right )-\cot \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+4 \cos \left (b x +a \right )^{4}-22 \cos \left (b x +a \right )^{2}+42 \cos \left (b x +a \right )-24\right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {d \tan \left (b x +a \right )}\, d \sqrt {2}}{24 b \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(288\)

Input:

int(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24/b*csc(b*x+a)*((21*cos(b*x+a)+21)*(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))+(-42*cos(b*x+a)-42)*(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))+4 
*cos(b*x+a)^4-22*cos(b*x+a)^2+42*cos(b*x+a)-24)*(-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)/(co 
s(b*x+a)+1)^2)^(1/2)*2^(1/2)
 

Fricas [F]

\[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

integral(-(d*cos(b*x + a)^2 - d)*sqrt(d*tan(b*x + a))*sin(b*x + a)*tan(b*x 
 + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**3*(d*tan(b*x+a))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} \sin \left (b x + a\right )^{3} \,d x } \] Input:

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*tan(b*x + a))^(3/2)*sin(b*x + a)^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sin(b*x+a)^3*(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
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\sin \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \] Input:

int(sin(a + b*x)^3*(d*tan(a + b*x))^(3/2),x)
 

Output:

int(sin(a + b*x)^3*(d*tan(a + b*x))^(3/2), x)
 

Reduce [F]

\[ \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx=\sqrt {d}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sin \left (b x +a \right )^{3} \tan \left (b x +a \right )d x \right ) d \] Input:

int(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x)
 

Output:

sqrt(d)*int(sqrt(tan(a + b*x))*sin(a + b*x)**3*tan(a + b*x),x)*d