Integrand size = 18, antiderivative size = 231 \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=-\frac {5 d (c+d x)^{3/2}}{16 b^2}+\frac {(c+d x)^{7/2}}{7 d}-\frac {15 d^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{128 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{128 b^{7/2}}-\frac {(c+d x)^{5/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}+\frac {15 d^2 \sqrt {c+d x} \sin (2 a+2 b x)}{64 b^3} \] Output:
-5/16*d*(d*x+c)^(3/2)/b^2+1/7*(d*x+c)^(7/2)/d-15/128*d^(5/2)*Pi^(1/2)*cos( 2*a-2*b*c/d)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))/b^(7/2)-15 /128*d^(5/2)*Pi^(1/2)*FresnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*s in(2*a-2*b*c/d)/b^(7/2)-1/2*(d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)/b+5/8*d*(d *x+c)^(3/2)*sin(b*x+a)^2/b^2+15/64*d^2*(d*x+c)^(1/2)*sin(2*b*x+2*a)/b^3
Result contains complex when optimal does not.
Time = 0.89 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.65 \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\frac {64 (c+d x)^4+\frac {7 \sqrt {2} d^4 e^{2 i \left (a-\frac {b c}{d}\right )} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {2 i b (c+d x)}{d}\right )}{b^4}+\frac {7 \sqrt {2} d^4 e^{-2 i \left (a-\frac {b c}{d}\right )} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {2 i b (c+d x)}{d}\right )}{b^4}}{448 d \sqrt {c+d x}} \] Input:
Integrate[(c + d*x)^(5/2)*Sin[a + b*x]^2,x]
Output:
(64*(c + d*x)^4 + (7*Sqrt[2]*d^4*E^((2*I)*(a - (b*c)/d))*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[7/2, ((-2*I)*b*(c + d*x))/d])/b^4 + (7*Sqrt[2]*d^4*Sqrt[(I *b*(c + d*x))/d]*Gamma[7/2, ((2*I)*b*(c + d*x))/d])/(b^4*E^((2*I)*(a - (b* c)/d))))/(448*d*Sqrt[c + d*x])
Time = 0.71 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3792, 17, 3042, 3793, 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.
| \(\displaystyle \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d x)^{5/2} \sin (a+b x)^2dx\) |
| \(\Big \downarrow \) 3792 |
| \(\displaystyle -\frac {15 d^2 \int \sqrt {c+d x} \sin ^2(a+b x)dx}{16 b^2}+\frac {1}{2} \int (c+d x)^{5/2}dx+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {15 d^2 \int \sqrt {c+d x} \sin ^2(a+b x)dx}{16 b^2}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {15 d^2 \int \sqrt {c+d x} \sin (a+b x)^2dx}{16 b^2}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle -\frac {15 d^2 \int \left (\frac {1}{2} \sqrt {c+d x}-\frac {1}{2} \sqrt {c+d x} \cos (2 a+2 b x)\right )dx}{16 b^2}+\frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5 d (c+d x)^{3/2} \sin ^2(a+b x)}{8 b^2}-\frac {15 d^2 \left (\frac {\sqrt {\pi } \sqrt {d} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}+\frac {\sqrt {\pi } \sqrt {d} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {c+d x} \sin (2 a+2 b x)}{4 b}+\frac {(c+d x)^{3/2}}{3 d}\right )}{16 b^2}-\frac {(c+d x)^{5/2} \sin (a+b x) \cos (a+b x)}{2 b}+\frac {(c+d x)^{7/2}}{7 d}\) |
Input:
Int[(c + d*x)^(5/2)*Sin[a + b*x]^2,x]
Output:
(c + d*x)^(7/2)/(7*d) - ((c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x])/(2*b) + (5*d*(c + d*x)^(3/2)*Sin[a + b*x]^2)/(8*b^2) - (15*d^2*((c + d*x)^(3/2)/ (3*d) + (Sqrt[d]*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(8*b^(3/2)) + (Sqrt[d]*Sqrt[Pi]*FresnelC[(2*S qrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(8*b^(3/2) ) - (Sqrt[c + d*x]*Sin[2*a + 2*b*x])/(4*b)))/(16*b^2)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 2*((n - 1)/n) Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 *m*((m - 1)/(f^2*n^2)) Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Time = 1.23 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d}\) | \(242\) |
| default | \(\frac {\frac {\left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 b c}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 a d -2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{4 b}}{d}\) | \(242\) |
Input:
int((d*x+c)^(5/2)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
2/d*(1/14*(d*x+c)^(7/2)-1/8/b*d*(d*x+c)^(5/2)*sin(2*b*(d*x+c)/d+2*(a*d-b*c )/d)+5/8/b*d*(-1/4/b*d*(d*x+c)^(3/2)*cos(2*b*(d*x+c)/d+2*(a*d-b*c)/d)+3/4/ b*d*(1/4/b*d*(d*x+c)^(1/2)*sin(2*b*(d*x+c)/d+2*(a*d-b*c)/d)-1/8/b*d*Pi^(1/ 2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+ c)^(1/2)/d)+sin(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^( 1/2)/d)))))
Time = 0.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.12 \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=-\frac {105 \, \pi d^{4} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 105 \, \pi d^{4} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, {\left (32 \, b^{4} d^{3} x^{3} + 96 \, b^{4} c d^{2} x^{2} + 32 \, b^{4} c^{3} + 70 \, b^{2} c d^{2} - 140 \, {\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{2} - 7 \, {\left (16 \, b^{3} d^{3} x^{2} + 32 \, b^{3} c d^{2} x + 16 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, {\left (48 \, b^{4} c^{2} d + 35 \, b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{896 \, b^{4} d} \] Input:
integrate((d*x+c)^(5/2)*sin(b*x+a)^2,x, algorithm="fricas")
Output:
-1/896*(105*pi*d^4*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt (d*x + c)*sqrt(b/(pi*d))) + 105*pi*d^4*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(d *x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 4*(32*b^4*d^3*x^3 + 96*b^4 *c*d^2*x^2 + 32*b^4*c^3 + 70*b^2*c*d^2 - 140*(b^2*d^3*x + b^2*c*d^2)*cos(b *x + a)^2 - 7*(16*b^3*d^3*x^2 + 32*b^3*c*d^2*x + 16*b^3*c^2*d - 15*b*d^3)* cos(b*x + a)*sin(b*x + a) + 2*(48*b^4*c^2*d + 35*b^2*d^3)*x)*sqrt(d*x + c) )/(b^4*d)
\[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\int \left (c + d x\right )^{\frac {5}{2}} \sin ^{2}{\left (a + b x \right )}\, dx \] Input:
integrate((d*x+c)**(5/2)*sin(b*x+a)**2,x)
Output:
Integral((c + d*x)**(5/2)*sin(a + b*x)**2, x)
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.28 \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\frac {\sqrt {2} {\left (\frac {512 \, \sqrt {2} {\left (d x + c\right )}^{\frac {7}{2}} b^{4}}{d} - 1120 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 105 \, {\left (-\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) + 105 \, {\left (\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d^{3} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right ) - 56 \, {\left (16 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 15 \, \sqrt {2} \sqrt {d x + c} b d^{2}\right )} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )\right )}}{7168 \, b^{4}} \] Input:
integrate((d*x+c)^(5/2)*sin(b*x+a)^2,x, algorithm="maxima")
Output:
1/7168*sqrt(2)*(512*sqrt(2)*(d*x + c)^(7/2)*b^4/d - 1120*sqrt(2)*(d*x + c) ^(3/2)*b^2*d*cos(2*((d*x + c)*b - b*c + a*d)/d) + 105*(-(I + 1)*4^(1/4)*sq rt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I - 1)*4^(1/4)*sqrt(pi )*d^3*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/ d)) + 105*((I - 1)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d) /d) - (I + 1)*4^(1/4)*sqrt(pi)*d^3*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))* erf(sqrt(d*x + c)*sqrt(-2*I*b/d)) - 56*(16*sqrt(2)*(d*x + c)^(5/2)*b^3 - 1 5*sqrt(2)*sqrt(d*x + c)*b*d^2)*sin(2*((d*x + c)*b - b*c + a*d)/d))/b^4
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 1310, normalized size of antiderivative = 5.67 \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^(5/2)*sin(b*x+a)^2,x, algorithm="giac")
Output:
-1/8960*(2240*(I*sqrt(pi)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2 *d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1 )) - I*sqrt(pi)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1) /d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)) - 4*s qrt(d*x + c))*c^3 - d^3*(256*(5*(d*x + c)^(7/2) - 21*(d*x + c)^(5/2)*c + 3 5*(d*x + c)^(3/2)*c^2 - 35*sqrt(d*x + c)*c^3)/d^3 - 35*(-I*sqrt(pi)*(64*b^ 3*c^3 - 48*I*b^2*c^2*d - 36*b*c*d^2 + 15*I*d^3)*d*erf(-I*sqrt(b*d)*sqrt(d* x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I *b*d/sqrt(b^2*d^2) + 1)*b^3) + 2*(-16*I*(d*x + c)^(5/2)*b^2*d + 48*I*(d*x + c)^(3/2)*b^2*c*d - 48*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d ^2 - 36*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^(-2*(-I*(d*x + c )*b + I*b*c - I*a*d)/d)/b^3)/d^3 - 35*(I*sqrt(pi)*(64*b^3*c^3 + 48*I*b^2*c ^2*d - 36*b*c*d^2 - 15*I*d^3)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt (b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^ 2) + 1)*b^3) + 2*(16*I*(d*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c* d + 48*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 - 36*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^(-2*(I*(d*x + c)*b - I*b*c + I*a* d)/d)/b^3)/d^3) + 560*(-3*I*sqrt(pi)*(4*b*c - I*d)*d*erf(-I*sqrt(b*d)*sqrt (d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d) *(I*b*d/sqrt(b^2*d^2) + 1)*b) + 3*I*sqrt(pi)*(4*b*c + I*d)*d*erf(I*sqrt...
Timed out. \[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2} \,d x \] Input:
int(sin(a + b*x)^2*(c + d*x)^(5/2),x)
Output:
int(sin(a + b*x)^2*(c + d*x)^(5/2), x)
\[ \int (c+d x)^{5/2} \sin ^2(a+b x) \, dx=\left (\int \sqrt {d x +c}\, \sin \left (b x +a \right )^{2} x^{2}d x \right ) d^{2}+2 \left (\int \sqrt {d x +c}\, \sin \left (b x +a \right )^{2} x d x \right ) c d +\left (\int \sqrt {d x +c}\, \sin \left (b x +a \right )^{2}d x \right ) c^{2} \] Input:
int((d*x+c)^(5/2)*sin(b*x+a)^2,x)
Output:
int(sqrt(c + d*x)*sin(a + b*x)**2*x**2,x)*d**2 + 2*int(sqrt(c + d*x)*sin(a + b*x)**2*x,x)*c*d + int(sqrt(c + d*x)*sin(a + b*x)**2,x)*c**2