\(\int (e+f x)^2 \sin (a+b \sqrt {c+d x}) \, dx\) [187]

Optimal result

Integrand size = 22, antiderivative size = 410 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=-\frac {240 f^2 \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^5 d^3}+\frac {24 f (d e-c f) \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^3}-\frac {2 (d e-c f)^2 \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b d^3}+\frac {40 f^2 (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^3}-\frac {4 f (d e-c f) (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^3}-\frac {2 f^2 (c+d x)^{5/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^3}+\frac {240 f^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^6 d^3}-\frac {24 f (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^3}+\frac {2 (d e-c f)^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^3}-\frac {120 f^2 (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^3}+\frac {12 f (d e-c f) (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^3}+\frac {10 f^2 (c+d x)^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^3} \] Output:

-240*f^2*(d*x+c)^(1/2)*cos(a+b*(d*x+c)^(1/2))/b^5/d^3+24*f*(-c*f+d*e)*(d*x 
+c)^(1/2)*cos(a+b*(d*x+c)^(1/2))/b^3/d^3-2*(-c*f+d*e)^2*(d*x+c)^(1/2)*cos( 
a+b*(d*x+c)^(1/2))/b/d^3+40*f^2*(d*x+c)^(3/2)*cos(a+b*(d*x+c)^(1/2))/b^3/d 
^3-4*f*(-c*f+d*e)*(d*x+c)^(3/2)*cos(a+b*(d*x+c)^(1/2))/b/d^3-2*f^2*(d*x+c) 
^(5/2)*cos(a+b*(d*x+c)^(1/2))/b/d^3+240*f^2*sin(a+b*(d*x+c)^(1/2))/b^6/d^3 
-24*f*(-c*f+d*e)*sin(a+b*(d*x+c)^(1/2))/b^4/d^3+2*(-c*f+d*e)^2*sin(a+b*(d* 
x+c)^(1/2))/b^2/d^3-120*f^2*(d*x+c)*sin(a+b*(d*x+c)^(1/2))/b^4/d^3+12*f*(- 
c*f+d*e)*(d*x+c)*sin(a+b*(d*x+c)^(1/2))/b^2/d^3+10*f^2*(d*x+c)^2*sin(a+b*( 
d*x+c)^(1/2))/b^2/d^3
 

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.34 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=\frac {-2 b \sqrt {c+d x} \left (120 f^2+b^4 d^2 (e+f x)^2-4 b^2 f (3 d e+2 c f+5 d f x)\right ) \cos \left (a+b \sqrt {c+d x}\right )+2 \left (120 f^2-12 b^2 f (4 c f+d (e+5 f x))+b^4 d (e+f x) (4 c f+d (e+5 f x))\right ) \sin \left (a+b \sqrt {c+d x}\right )}{b^6 d^3} \] Input:

Integrate[(e + f*x)^2*Sin[a + b*Sqrt[c + d*x]],x]
 

Output:

(-2*b*Sqrt[c + d*x]*(120*f^2 + b^4*d^2*(e + f*x)^2 - 4*b^2*f*(3*d*e + 2*c* 
f + 5*d*f*x))*Cos[a + b*Sqrt[c + d*x]] + 2*(120*f^2 - 12*b^2*f*(4*c*f + d* 
(e + 5*f*x)) + b^4*d*(e + f*x)*(4*c*f + d*(e + 5*f*x)))*Sin[a + b*Sqrt[c + 
 d*x]])/(b^6*d^3)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3912, 2009}

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.

\(\displaystyle \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx\)

\(\Big \downarrow \) 3912

\(\displaystyle \frac {2 \int \left (\frac {f^2 \sin \left (a+b \sqrt {c+d x}\right ) (c+d x)^{5/2}}{d^2}+\frac {2 f (d e-c f) \sin \left (a+b \sqrt {c+d x}\right ) (c+d x)^{3/2}}{d^2}+\frac {(d e-c f)^2 \sin \left (a+b \sqrt {c+d x}\right ) \sqrt {c+d x}}{d^2}\right )d\sqrt {c+d x}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {120 f^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^6 d^2}-\frac {120 f^2 \sqrt {c+d x} \cos \left (a+b \sqrt {c+d x}\right )}{b^5 d^2}-\frac {12 f (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}-\frac {60 f^2 (c+d x) \sin \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}+\frac {12 f \sqrt {c+d x} (d e-c f) \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}+\frac {20 f^2 (c+d x)^{3/2} \cos \left (a+b \sqrt {c+d x}\right )}{b^3 d^2}+\frac {6 f (c+d x) (d e-c f) \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {(d e-c f)^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}+\frac {5 f^2 (c+d x)^2 \sin \left (a+b \sqrt {c+d x}\right )}{b^2 d^2}-\frac {2 f (c+d x)^{3/2} (d e-c f) \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {\sqrt {c+d x} (d e-c f)^2 \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}-\frac {f^2 (c+d x)^{5/2} \cos \left (a+b \sqrt {c+d x}\right )}{b d^2}\right )}{d}\)

Input:

Int[(e + f*x)^2*Sin[a + b*Sqrt[c + d*x]],x]
 

Output:

(2*((-120*f^2*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b^5*d^2) + (12*f*(d 
*e - c*f)*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b^3*d^2) - ((d*e - c*f) 
^2*Sqrt[c + d*x]*Cos[a + b*Sqrt[c + d*x]])/(b*d^2) + (20*f^2*(c + d*x)^(3/ 
2)*Cos[a + b*Sqrt[c + d*x]])/(b^3*d^2) - (2*f*(d*e - c*f)*(c + d*x)^(3/2)* 
Cos[a + b*Sqrt[c + d*x]])/(b*d^2) - (f^2*(c + d*x)^(5/2)*Cos[a + b*Sqrt[c 
+ d*x]])/(b*d^2) + (120*f^2*Sin[a + b*Sqrt[c + d*x]])/(b^6*d^2) - (12*f*(d 
*e - c*f)*Sin[a + b*Sqrt[c + d*x]])/(b^4*d^2) + ((d*e - c*f)^2*Sin[a + b*S 
qrt[c + d*x]])/(b^2*d^2) - (60*f^2*(c + d*x)*Sin[a + b*Sqrt[c + d*x]])/(b^ 
4*d^2) + (6*f*(d*e - c*f)*(c + d*x)*Sin[a + b*Sqrt[c + d*x]])/(b^2*d^2) + 
(5*f^2*(c + d*x)^2*Sin[a + b*Sqrt[c + d*x]])/(b^2*d^2)))/d
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3912

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Simp[1/(n*f)   Subst[Int[ExpandIntegra 
nd[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - e*(h/f) + h*(x^(1/n)/f))^m, x], 
 x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p 
, 0] && IntegerQ[1/n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1160\) vs. \(2(374)=748\).

Time = 1.23 (sec) , antiderivative size = 1161, normalized size of antiderivative = 2.83

method	result	size
parts	\(\text {Expression too large to display}\)	\(1161\)
derivativedivides	\(\text {Expression too large to display}\)	\(1246\)
default	\(\text {Expression too large to display}\)	\(1246\)

Input:

int((f*x+e)^2*sin(a+(d*x+c)^(1/2)*b),x,method=_RETURNVERBOSE)
 

Output:

-2/d/b*(d*x+c)^(1/2)*cos(a+(d*x+c)^(1/2)*b)*f^2*x^2-4/d/b*(d*x+c)^(1/2)*co 
s(a+(d*x+c)^(1/2)*b)*e*f*x-2/d/b*(d*x+c)^(1/2)*cos(a+(d*x+c)^(1/2)*b)*e^2+ 
2/d/b^2*sin(a+(d*x+c)^(1/2)*b)*f^2*x^2+4/d/b^2*sin(a+(d*x+c)^(1/2)*b)*e*f* 
x+2/d/b^2*sin(a+(d*x+c)^(1/2)*b)*e^2-8/d^3/b^4*f*(3/b^2*a^2*f*(sin(a+(d*x+ 
c)^(1/2)*b)-(a+(d*x+c)^(1/2)*b)*cos(a+(d*x+c)^(1/2)*b))-3/b^2*a*f*(-(a+(d* 
x+c)^(1/2)*b)^2*cos(a+(d*x+c)^(1/2)*b)+2*cos(a+(d*x+c)^(1/2)*b)+2*(a+(d*x+ 
c)^(1/2)*b)*sin(a+(d*x+c)^(1/2)*b))-6/b^2*a^2*f*((a+(d*x+c)^(1/2)*b)^2*sin 
(a+(d*x+c)^(1/2)*b)-2*sin(a+(d*x+c)^(1/2)*b)+2*(a+(d*x+c)^(1/2)*b)*cos(a+( 
d*x+c)^(1/2)*b))+4/b^2*a^3*f*(cos(a+(d*x+c)^(1/2)*b)+(a+(d*x+c)^(1/2)*b)*s 
in(a+(d*x+c)^(1/2)*b))+4/b^2*a*f*((a+(d*x+c)^(1/2)*b)^3*sin(a+(d*x+c)^(1/2 
)*b)+3*(a+(d*x+c)^(1/2)*b)^2*cos(a+(d*x+c)^(1/2)*b)-6*cos(a+(d*x+c)^(1/2)* 
b)-6*(a+(d*x+c)^(1/2)*b)*sin(a+(d*x+c)^(1/2)*b))-c*f*(sin(a+(d*x+c)^(1/2)* 
b)-(a+(d*x+c)^(1/2)*b)*cos(a+(d*x+c)^(1/2)*b))-c*f*a*cos(a+(d*x+c)^(1/2)*b 
)+d*e*(sin(a+(d*x+c)^(1/2)*b)-(a+(d*x+c)^(1/2)*b)*cos(a+(d*x+c)^(1/2)*b))+ 
d*e*a*cos(a+(d*x+c)^(1/2)*b)+c*f*((a+(d*x+c)^(1/2)*b)^2*sin(a+(d*x+c)^(1/2 
)*b)-2*sin(a+(d*x+c)^(1/2)*b)+2*(a+(d*x+c)^(1/2)*b)*cos(a+(d*x+c)^(1/2)*b) 
)-d*e*((a+(d*x+c)^(1/2)*b)^2*sin(a+(d*x+c)^(1/2)*b)-2*sin(a+(d*x+c)^(1/2)* 
b)+2*(a+(d*x+c)^(1/2)*b)*cos(a+(d*x+c)^(1/2)*b))+a^2*c*f*sin(a+(d*x+c)^(1/ 
2)*b)-a^2*d*e*sin(a+(d*x+c)^(1/2)*b)-2*c*f*a*(cos(a+(d*x+c)^(1/2)*b)+(a+(d 
*x+c)^(1/2)*b)*sin(a+(d*x+c)^(1/2)*b))+2*d*e*a*(cos(a+(d*x+c)^(1/2)*b)+...
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.48 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=-\frac {2 \, {\left ({\left (b^{5} d^{2} f^{2} x^{2} + b^{5} d^{2} e^{2} - 12 \, b^{3} d e f - 8 \, {\left (b^{3} c - 15 \, b\right )} f^{2} + 2 \, {\left (b^{5} d^{2} e f - 10 \, b^{3} d f^{2}\right )} x\right )} \sqrt {d x + c} \cos \left (\sqrt {d x + c} b + a\right ) - {\left (5 \, b^{4} d^{2} f^{2} x^{2} + b^{4} d^{2} e^{2} + 4 \, {\left (b^{4} c - 3 \, b^{2}\right )} d e f - 24 \, {\left (2 \, b^{2} c - 5\right )} f^{2} + 2 \, {\left (3 \, b^{4} d^{2} e f + 2 \, {\left (b^{4} c - 15 \, b^{2}\right )} d f^{2}\right )} x\right )} \sin \left (\sqrt {d x + c} b + a\right )\right )}}{b^{6} d^{3}} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/2)),x, algorithm="fricas")
 

Output:

-2*((b^5*d^2*f^2*x^2 + b^5*d^2*e^2 - 12*b^3*d*e*f - 8*(b^3*c - 15*b)*f^2 + 
 2*(b^5*d^2*e*f - 10*b^3*d*f^2)*x)*sqrt(d*x + c)*cos(sqrt(d*x + c)*b + a) 
- (5*b^4*d^2*f^2*x^2 + b^4*d^2*e^2 + 4*(b^4*c - 3*b^2)*d*e*f - 24*(2*b^2*c 
 - 5)*f^2 + 2*(3*b^4*d^2*e*f + 2*(b^4*c - 15*b^2)*d*f^2)*x)*sin(sqrt(d*x + 
 c)*b + a))/(b^6*d^3)
 

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.29 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=\begin {cases} \left (e^{2} x + e f x^{2} + \frac {f^{2} x^{3}}{3}\right ) \sin {\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\\left (e^{2} x + e f x^{2} + \frac {f^{2} x^{3}}{3}\right ) \sin {\left (a + b \sqrt {c} \right )} & \text {for}\: d = 0 \\- \frac {2 e^{2} \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} - \frac {4 e f x \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} - \frac {2 f^{2} x^{2} \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b d} + \frac {8 c e f \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d^{2}} + \frac {8 c f^{2} x \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d^{2}} + \frac {2 e^{2} \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {12 e f x \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {10 f^{2} x^{2} \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{2} d} + \frac {16 c f^{2} \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b^{3} d^{3}} + \frac {24 e f \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b^{3} d^{2}} + \frac {40 f^{2} x \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b^{3} d^{2}} - \frac {96 c f^{2} \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{4} d^{3}} - \frac {24 e f \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{4} d^{2}} - \frac {120 f^{2} x \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{4} d^{2}} - \frac {240 f^{2} \sqrt {c + d x} \cos {\left (a + b \sqrt {c + d x} \right )}}{b^{5} d^{3}} + \frac {240 f^{2} \sin {\left (a + b \sqrt {c + d x} \right )}}{b^{6} d^{3}} & \text {otherwise} \end {cases} \] Input:

integrate((f*x+e)**2*sin(a+b*(d*x+c)**(1/2)),x)
                                                                                    
                                                                                    
 

Output:

Piecewise(((e**2*x + e*f*x**2 + f**2*x**3/3)*sin(a), Eq(b, 0) & (Eq(b, 0) 
| Eq(d, 0))), ((e**2*x + e*f*x**2 + f**2*x**3/3)*sin(a + b*sqrt(c)), Eq(d, 
 0)), (-2*e**2*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b*d) - 4*e*f*x*sqrt 
(c + d*x)*cos(a + b*sqrt(c + d*x))/(b*d) - 2*f**2*x**2*sqrt(c + d*x)*cos(a 
 + b*sqrt(c + d*x))/(b*d) + 8*c*e*f*sin(a + b*sqrt(c + d*x))/(b**2*d**2) + 
 8*c*f**2*x*sin(a + b*sqrt(c + d*x))/(b**2*d**2) + 2*e**2*sin(a + b*sqrt(c 
 + d*x))/(b**2*d) + 12*e*f*x*sin(a + b*sqrt(c + d*x))/(b**2*d) + 10*f**2*x 
**2*sin(a + b*sqrt(c + d*x))/(b**2*d) + 16*c*f**2*sqrt(c + d*x)*cos(a + b* 
sqrt(c + d*x))/(b**3*d**3) + 24*e*f*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x)) 
/(b**3*d**2) + 40*f**2*x*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b**3*d**2 
) - 96*c*f**2*sin(a + b*sqrt(c + d*x))/(b**4*d**3) - 24*e*f*sin(a + b*sqrt 
(c + d*x))/(b**4*d**2) - 120*f**2*x*sin(a + b*sqrt(c + d*x))/(b**4*d**2) - 
 240*f**2*sqrt(c + d*x)*cos(a + b*sqrt(c + d*x))/(b**5*d**3) + 240*f**2*si 
n(a + b*sqrt(c + d*x))/(b**6*d**3), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1101 vs. \(2 (374) = 748\).

Time = 0.09 (sec) , antiderivative size = 1101, normalized size of antiderivative = 2.69 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")
 

Output:

2*(a*e^2*cos(sqrt(d*x + c)*b + a) - 2*a*c*e*f*cos(sqrt(d*x + c)*b + a)/d + 
 a*c^2*f^2*cos(sqrt(d*x + c)*b + a)/d^2 - ((sqrt(d*x + c)*b + a)*cos(sqrt( 
d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))*e^2 + 2*((sqrt(d*x + c)*b + a) 
*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))*c*e*f/d - ((sqrt(d*x 
 + c)*b + a)*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))*c^2*f^2/ 
d^2 + 2*a^3*e*f*cos(sqrt(d*x + c)*b + a)/(b^2*d) - 2*a^3*c*f^2*cos(sqrt(d* 
x + c)*b + a)/(b^2*d^2) - 6*((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*b + a 
) - sin(sqrt(d*x + c)*b + a))*a^2*e*f/(b^2*d) + 6*((sqrt(d*x + c)*b + a)*c 
os(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))*a^2*c*f^2/(b^2*d^2) + 
a^5*f^2*cos(sqrt(d*x + c)*b + a)/(b^4*d^2) + 6*(((sqrt(d*x + c)*b + a)^2 - 
 2)*cos(sqrt(d*x + c)*b + a) - 2*(sqrt(d*x + c)*b + a)*sin(sqrt(d*x + c)*b 
 + a))*a*e*f/(b^2*d) - 5*((sqrt(d*x + c)*b + a)*cos(sqrt(d*x + c)*b + a) - 
 sin(sqrt(d*x + c)*b + a))*a^4*f^2/(b^4*d^2) - 6*(((sqrt(d*x + c)*b + a)^2 
 - 2)*cos(sqrt(d*x + c)*b + a) - 2*(sqrt(d*x + c)*b + a)*sin(sqrt(d*x + c) 
*b + a))*a*c*f^2/(b^2*d^2) - 2*(((sqrt(d*x + c)*b + a)^3 - 6*sqrt(d*x + c) 
*b - 6*a)*cos(sqrt(d*x + c)*b + a) - 3*((sqrt(d*x + c)*b + a)^2 - 2)*sin(s 
qrt(d*x + c)*b + a))*e*f/(b^2*d) + 10*(((sqrt(d*x + c)*b + a)^2 - 2)*cos(s 
qrt(d*x + c)*b + a) - 2*(sqrt(d*x + c)*b + a)*sin(sqrt(d*x + c)*b + a))*a^ 
3*f^2/(b^4*d^2) + 2*(((sqrt(d*x + c)*b + a)^3 - 6*sqrt(d*x + c)*b - 6*a)*c 
os(sqrt(d*x + c)*b + a) - 3*((sqrt(d*x + c)*b + a)^2 - 2)*sin(sqrt(d*x ...
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 698, normalized size of antiderivative = 1.70 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx =\text {Too large to display} \] Input:

integrate((f*x+e)^2*sin(a+b*(d*x+c)^(1/2)),x, algorithm="giac")
 

Output:

-2*((sqrt(d*x + c)*b*cos(sqrt(d*x + c)*b + a) - sin(sqrt(d*x + c)*b + a))* 
e^2/b - 2*e*f*(((sqrt(d*x + c)*b + a)*b^2*c - a*b^2*c - (sqrt(d*x + c)*b + 
 a)^3 + 3*(sqrt(d*x + c)*b + a)^2*a - 3*(sqrt(d*x + c)*b + a)*a^2 + a^3 + 
6*sqrt(d*x + c)*b)*cos(sqrt(d*x + c)*b + a)/b^2 - (b^2*c - 3*(sqrt(d*x + c 
)*b + a)^2 + 6*(sqrt(d*x + c)*b + a)*a - 3*a^2 + 6)*sin(sqrt(d*x + c)*b + 
a)/b^2)/(b*d) + f^2*(((sqrt(d*x + c)*b + a)*b^4*c^2 - a*b^4*c^2 - 2*(sqrt( 
d*x + c)*b + a)^3*b^2*c + 6*(sqrt(d*x + c)*b + a)^2*a*b^2*c - 6*(sqrt(d*x 
+ c)*b + a)*a^2*b^2*c + 2*a^3*b^2*c + (sqrt(d*x + c)*b + a)^5 - 5*(sqrt(d* 
x + c)*b + a)^4*a + 10*(sqrt(d*x + c)*b + a)^3*a^2 - 10*(sqrt(d*x + c)*b + 
 a)^2*a^3 + 5*(sqrt(d*x + c)*b + a)*a^4 - a^5 + 12*(sqrt(d*x + c)*b + a)*b 
^2*c - 12*a*b^2*c - 20*(sqrt(d*x + c)*b + a)^3 + 60*(sqrt(d*x + c)*b + a)^ 
2*a - 60*(sqrt(d*x + c)*b + a)*a^2 + 20*a^3 + 120*sqrt(d*x + c)*b)*cos(sqr 
t(d*x + c)*b + a)/b^4 - (b^4*c^2 - 6*(sqrt(d*x + c)*b + a)^2*b^2*c + 12*(s 
qrt(d*x + c)*b + a)*a*b^2*c - 6*a^2*b^2*c + 5*(sqrt(d*x + c)*b + a)^4 - 20 
*(sqrt(d*x + c)*b + a)^3*a + 30*(sqrt(d*x + c)*b + a)^2*a^2 - 20*(sqrt(d*x 
 + c)*b + a)*a^3 + 5*a^4 + 12*b^2*c - 60*(sqrt(d*x + c)*b + a)^2 + 120*(sq 
rt(d*x + c)*b + a)*a - 60*a^2 + 120)*sin(sqrt(d*x + c)*b + a)/b^4)/(b*d^2) 
)/(b*d)
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=\int \sin \left (a+b\,\sqrt {c+d\,x}\right )\,{\left (e+f\,x\right )}^2 \,d x \] Input:

int(sin(a + b*(c + d*x)^(1/2))*(e + f*x)^2,x)
 

Output:

int(sin(a + b*(c + d*x)^(1/2))*(e + f*x)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.93 \[ \int (e+f x)^2 \sin \left (a+b \sqrt {c+d x}\right ) \, dx=\frac {-2 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{5} d^{2} e^{2}-4 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{5} d^{2} e f x -2 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{5} d^{2} f^{2} x^{2}+16 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{3} c \,f^{2}+24 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{3} d e f +40 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b^{3} d \,f^{2} x -240 \sqrt {d x +c}\, \cos \left (\sqrt {d x +c}\, b +a \right ) b \,f^{2}+8 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{4} c d e f +8 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{4} c d \,f^{2} x +2 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{4} d^{2} e^{2}+12 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{4} d^{2} e f x +10 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{4} d^{2} f^{2} x^{2}-96 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{2} c \,f^{2}-24 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{2} d e f -120 \sin \left (\sqrt {d x +c}\, b +a \right ) b^{2} d \,f^{2} x +240 \sin \left (\sqrt {d x +c}\, b +a \right ) f^{2}}{b^{6} d^{3}} \] Input:

int((f*x+e)^2*sin(a+b*(d*x+c)^(1/2)),x)
 

Output:

(2*( - sqrt(c + d*x)*cos(sqrt(c + d*x)*b + a)*b**5*d**2*e**2 - 2*sqrt(c + 
d*x)*cos(sqrt(c + d*x)*b + a)*b**5*d**2*e*f*x - sqrt(c + d*x)*cos(sqrt(c + 
 d*x)*b + a)*b**5*d**2*f**2*x**2 + 8*sqrt(c + d*x)*cos(sqrt(c + d*x)*b + a 
)*b**3*c*f**2 + 12*sqrt(c + d*x)*cos(sqrt(c + d*x)*b + a)*b**3*d*e*f + 20* 
sqrt(c + d*x)*cos(sqrt(c + d*x)*b + a)*b**3*d*f**2*x - 120*sqrt(c + d*x)*c 
os(sqrt(c + d*x)*b + a)*b*f**2 + 4*sin(sqrt(c + d*x)*b + a)*b**4*c*d*e*f + 
 4*sin(sqrt(c + d*x)*b + a)*b**4*c*d*f**2*x + sin(sqrt(c + d*x)*b + a)*b** 
4*d**2*e**2 + 6*sin(sqrt(c + d*x)*b + a)*b**4*d**2*e*f*x + 5*sin(sqrt(c + 
d*x)*b + a)*b**4*d**2*f**2*x**2 - 48*sin(sqrt(c + d*x)*b + a)*b**2*c*f**2 
- 12*sin(sqrt(c + d*x)*b + a)*b**2*d*e*f - 60*sin(sqrt(c + d*x)*b + a)*b** 
2*d*f**2*x + 120*sin(sqrt(c + d*x)*b + a)*f**2))/(b**6*d**3)