\(\int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx\) [59]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {5 d \sin (a+b x)}{6 b \sqrt {d \tan (a+b x)}}-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}+\frac {5 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{12 b} \] Output:

-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)*InverseJacobiAM(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
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 11.61 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.32 \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=-\frac {\cos (2 (a+b x)) \sec (a+b x) \left (-5 \sqrt [4]{-1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right ) \sec ^2(a+b x)+(-6+\cos (2 (a+b x))) \sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)}\right ) \sqrt {d \tan (a+b x)}}{6 b \sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)} \left (-1+\tan ^2(a+b x)\right )} \] Input:

Integrate[Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]],x]
 

Output:

-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))
 

Rubi [A] (verified)

Time = 0.56 (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

\(\displaystyle \frac {5}{6} \left (\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{2 b}-\frac {d \sin (a+b x)}{b \sqrt {d \tan (a+b x)}}\right )-\frac {d \sin ^3(a+b x)}{3 b \sqrt {d \tan (a+b x)}}\)

Input:

Int[Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]],x]
 

Output:

-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
 

Defintions of rubi rules used

rule 3042

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

rule 3053

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]
 

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 3120

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]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.19 (sec) , antiderivative size = 1585, normalized size of antiderivative = 15.10

method	result	size
default	\(\text {Expression too large to display}\)	\(1585\)

Input:

int(sin(b*x+a)^3*(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/96/b*(d*tan(b*x+a))^(1/2)*(2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a) 
+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)-2*sin(b*x+a)*(-2*sin( 
b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+ 
a)+2)/(-1+cos(b*x+a)))*(6*cos(b*x+a)^2+6*cos(b*x+a)-3*cot(b*x+a)-3*csc(b*x 
+a))+(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b 
*x+a)-2*cot(b*x+a)-2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2 
)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+a)+2)/(-1+cos(b*x+a)))*(-12*cos(b* 
x+a)^2-12*cos(b*x+a))+2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^ 
(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-2*sin(b*x+a)* 
cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+a)+2)/( 
-1+cos(b*x+a)))*(-6*cos(b*x+a)^2-6*cos(b*x+a)+3*cot(b*x+a)+3*csc(b*x+a))+( 
-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)- 
2*cot(b*x+a)+2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2 
)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+a)+2)/(-1+cos(b*x+a)))*(12*cos(b*x+a)^2+ 
12*cos(b*x+a))+2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*a 
rctan((-sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b 
*x+a)-1)/(-1+cos(b*x+a)))*(12*cos(b*x+a)^2+12*cos(b*x+a)-6*cot(b*x+a)-6*cs 
c(b*x+a))+arctan((-sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^ 
(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1 
)^2)^(1/2)*(-24*cos(b*x+a)^2-24*cos(b*x+a))+2^(1/2)*(-2*sin(b*x+a)*cos(...
 

Fricas [F]

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

Sympy [F(-1)]

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

integrate(sin(b*x+a)**3*(d*tan(b*x+a))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(1/2),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. \[ \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx=\int {\sin \left (a+b\,x\right )}^3\,\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )} \,d x \] 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)
 

Reduce [F]

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