Integrand size = 22, antiderivative size = 382 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=-\frac {4 f (d e-c f) \sqrt {c+d x} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b d^3}-\frac {2 e^{i a} f (d e-c f) \sqrt {c+d x} \Gamma \left (\frac {1}{3},-i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{-i b (c+d x)^{3/2}}}-\frac {2 e^{-i a} f (d e-c f) \sqrt {c+d x} \Gamma \left (\frac {1}{3},i b (c+d x)^{3/2}\right )}{9 b d^3 \sqrt [3]{i b (c+d x)^{3/2}}}+\frac {i e^{i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{3 d^3 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{3 d^3 \left (i b (c+d x)^{3/2}\right )^{2/3}}+\frac {2 f^2 \sin \left (a+b (c+d x)^{3/2}\right )}{3 b^2 d^3} \]
-2/3*f^2*(d*x+c)^(3/2)*cos(a+b*(d*x+c)^(3/2))/b/d^3+1/3*I*exp(I*a)*(-c*f+d *e)^2*(d*x+c)*GAMMA(2/3,-I*b*(d*x+c)^(3/2))/d^3/(-I*b*(d*x+c)^(3/2))^(2/3) -1/3*I*(-c*f+d*e)^2*(d*x+c)*GAMMA(2/3,I*b*(d*x+c)^(3/2))/d^3/exp(I*a)/(I*b *(d*x+c)^(3/2))^(2/3)+2/3*f^2*sin(a+b*(d*x+c)^(3/2))/b^2/d^3-4/3*f*(-c*f+d *e)*cos(a+b*(d*x+c)^(3/2))*(d*x+c)^(1/2)/b/d^3-2/9*exp(I*a)*f*(-c*f+d*e)*G AMMA(1/3,-I*b*(d*x+c)^(3/2))*(d*x+c)^(1/2)/b/d^3/(-I*b*(d*x+c)^(3/2))^(1/3 )-2/9*f*(-c*f+d*e)*GAMMA(1/3,I*b*(d*x+c)^(3/2))*(d*x+c)^(1/2)/b/d^3/exp(I* a)/(I*b*(d*x+c)^(3/2))^(1/3)
Time = 2.11 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.10 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=-\frac {i \left ((\cos (a)+i \sin (a)) \left (\frac {f^2 \cos \left (b (c+d x)^{3/2}\right )}{b^2}-\frac {(d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{\left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {2 f (d e-c f) (c+d x)^2 \Gamma \left (\frac {4}{3},-i b (c+d x)^{3/2}\right )}{\left (-i b (c+d x)^{3/2}\right )^{4/3}}+\frac {i f^2 \sin \left (b (c+d x)^{3/2}\right )}{b^2}+\frac {f^2 (c+d x)^{3/2} \left (-i \cos \left (b (c+d x)^{3/2}\right )+\sin \left (b (c+d x)^{3/2}\right )\right )}{b}\right )-(\cos (a)-i \sin (a)) \left (\frac {f^2 \cos \left (b (c+d x)^{3/2}\right )}{b^2}-\frac {(d e-c f)^2 (c+d x) \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{\left (i b (c+d x)^{3/2}\right )^{2/3}}-\frac {2 f (d e-c f) (c+d x)^2 \Gamma \left (\frac {4}{3},i b (c+d x)^{3/2}\right )}{\left (i b (c+d x)^{3/2}\right )^{4/3}}-\frac {i f^2 \sin \left (b (c+d x)^{3/2}\right )}{b^2}+\frac {f^2 (c+d x)^{3/2} \left (i \cos \left (b (c+d x)^{3/2}\right )+\sin \left (b (c+d x)^{3/2}\right )\right )}{b}\right )\right )}{3 d^3} \]
((-1/3*I)*((Cos[a] + I*Sin[a])*((f^2*Cos[b*(c + d*x)^(3/2)])/b^2 - ((d*e - c*f)^2*(c + d*x)*Gamma[2/3, (-I)*b*(c + d*x)^(3/2)])/((-I)*b*(c + d*x)^(3 /2))^(2/3) - (2*f*(d*e - c*f)*(c + d*x)^2*Gamma[4/3, (-I)*b*(c + d*x)^(3/2 )])/((-I)*b*(c + d*x)^(3/2))^(4/3) + (I*f^2*Sin[b*(c + d*x)^(3/2)])/b^2 + (f^2*(c + d*x)^(3/2)*((-I)*Cos[b*(c + d*x)^(3/2)] + Sin[b*(c + d*x)^(3/2)] ))/b) - (Cos[a] - I*Sin[a])*((f^2*Cos[b*(c + d*x)^(3/2)])/b^2 - ((d*e - c* f)^2*(c + d*x)*Gamma[2/3, I*b*(c + d*x)^(3/2)])/(I*b*(c + d*x)^(3/2))^(2/3 ) - (2*f*(d*e - c*f)*(c + d*x)^2*Gamma[4/3, I*b*(c + d*x)^(3/2)])/(I*b*(c + d*x)^(3/2))^(4/3) - (I*f^2*Sin[b*(c + d*x)^(3/2)])/b^2 + (f^2*(c + d*x)^ (3/2)*(I*Cos[b*(c + d*x)^(3/2)] + Sin[b*(c + d*x)^(3/2)]))/b)))/d^3
Time = 0.43 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3914, 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.
| \(\displaystyle \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx\) |
| \(\Big \downarrow \) 3914 |
| \(\displaystyle \frac {2 \int \left (f^2 \sin \left (b (c+d x)^{3/2}+a\right ) (c+d x)^{5/2}+2 f (d e-c f) \sin \left (b (c+d x)^{3/2}+a\right ) (c+d x)^{3/2}+(d e-c f)^2 \sin \left (b (c+d x)^{3/2}+a\right ) \sqrt {c+d x}\right )d\sqrt {c+d x}}{d^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {f^2 \sin \left (a+b (c+d x)^{3/2}\right )}{3 b^2}-\frac {2 f \sqrt {c+d x} (d e-c f) \cos \left (a+b (c+d x)^{3/2}\right )}{3 b}-\frac {e^{i a} f \sqrt {c+d x} (d e-c f) \Gamma \left (\frac {1}{3},-i b (c+d x)^{3/2}\right )}{9 b \sqrt [3]{-i b (c+d x)^{3/2}}}-\frac {e^{-i a} f \sqrt {c+d x} (d e-c f) \Gamma \left (\frac {1}{3},i b (c+d x)^{3/2}\right )}{9 b \sqrt [3]{i b (c+d x)^{3/2}}}+\frac {i e^{i a} (c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},-i b (c+d x)^{3/2}\right )}{6 \left (-i b (c+d x)^{3/2}\right )^{2/3}}-\frac {i e^{-i a} (c+d x) (d e-c f)^2 \Gamma \left (\frac {2}{3},i b (c+d x)^{3/2}\right )}{6 \left (i b (c+d x)^{3/2}\right )^{2/3}}-\frac {f^2 (c+d x)^{3/2} \cos \left (a+b (c+d x)^{3/2}\right )}{3 b}\right )}{d^3}\) |
(2*((-2*f*(d*e - c*f)*Sqrt[c + d*x]*Cos[a + b*(c + d*x)^(3/2)])/(3*b) - (f ^2*(c + d*x)^(3/2)*Cos[a + b*(c + d*x)^(3/2)])/(3*b) - (E^(I*a)*f*(d*e - c *f)*Sqrt[c + d*x]*Gamma[1/3, (-I)*b*(c + d*x)^(3/2)])/(9*b*((-I)*b*(c + d* x)^(3/2))^(1/3)) - (f*(d*e - c*f)*Sqrt[c + d*x]*Gamma[1/3, I*b*(c + d*x)^( 3/2)])/(9*b*E^(I*a)*(I*b*(c + d*x)^(3/2))^(1/3)) + ((I/6)*E^(I*a)*(d*e - c *f)^2*(c + d*x)*Gamma[2/3, (-I)*b*(c + d*x)^(3/2)])/((-I)*b*(c + d*x)^(3/2 ))^(2/3) - ((I/6)*(d*e - c*f)^2*(c + d*x)*Gamma[2/3, I*b*(c + d*x)^(3/2)]) /(E^(I*a)*(I*b*(c + d*x)^(3/2))^(2/3)) + (f^2*Sin[a + b*(c + d*x)^(3/2)])/ (3*b^2)))/d^3
3.2.92.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f _.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat or[n], 1]}, Simp[k/f^(m + 1) Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x ^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x ]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
\[\int \left (f x +e \right )^{2} \sin \left (a +b \left (d x +c \right )^{\frac {3}{2}}\right )d x\]
Time = 0.12 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.96 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=\frac {6 \, f^{2} \sin \left ({\left (b d x + b c\right )} \sqrt {d x + c} + a\right ) - 2 \, {\left ({\left (-i \, d e f + i \, c f^{2}\right )} \cos \left (a\right ) - {\left (d e f - c f^{2}\right )} \sin \left (a\right )\right )} \left (i \, b\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, {\left (i \, b d x + i \, b c\right )} \sqrt {d x + c}\right ) - 2 \, {\left ({\left (i \, d e f - i \, c f^{2}\right )} \cos \left (a\right ) - {\left (d e f - c f^{2}\right )} \sin \left (a\right )\right )} \left (-i \, b\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, {\left (-i \, b d x - i \, b c\right )} \sqrt {d x + c}\right ) - 6 \, {\left (b d f^{2} x + 2 \, b d e f - b c f^{2}\right )} \sqrt {d x + c} \cos \left ({\left (b d x + b c\right )} \sqrt {d x + c} + a\right ) - 3 \, {\left ({\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \cos \left (a\right ) + {\left (-i \, b d^{2} e^{2} + 2 i \, b c d e f - i \, b c^{2} f^{2}\right )} \sin \left (a\right )\right )} \left (i \, b\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, {\left (i \, b d x + i \, b c\right )} \sqrt {d x + c}\right ) - 3 \, {\left ({\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \cos \left (a\right ) + {\left (i \, b d^{2} e^{2} - 2 i \, b c d e f + i \, b c^{2} f^{2}\right )} \sin \left (a\right )\right )} \left (-i \, b\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, {\left (-i \, b d x - i \, b c\right )} \sqrt {d x + c}\right )}{9 \, b^{2} d^{3}} \]
1/9*(6*f^2*sin((b*d*x + b*c)*sqrt(d*x + c) + a) - 2*((-I*d*e*f + I*c*f^2)* cos(a) - (d*e*f - c*f^2)*sin(a))*(I*b)^(2/3)*gamma(1/3, (I*b*d*x + I*b*c)* sqrt(d*x + c)) - 2*((I*d*e*f - I*c*f^2)*cos(a) - (d*e*f - c*f^2)*sin(a))*( -I*b)^(2/3)*gamma(1/3, (-I*b*d*x - I*b*c)*sqrt(d*x + c)) - 6*(b*d*f^2*x + 2*b*d*e*f - b*c*f^2)*sqrt(d*x + c)*cos((b*d*x + b*c)*sqrt(d*x + c) + a) - 3*((b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*cos(a) + (-I*b*d^2*e^2 + 2*I*b*c* d*e*f - I*b*c^2*f^2)*sin(a))*(I*b)^(1/3)*gamma(2/3, (I*b*d*x + I*b*c)*sqrt (d*x + c)) - 3*((b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*cos(a) + (I*b*d^2*e^ 2 - 2*I*b*c*d*e*f + I*b*c^2*f^2)*sin(a))*(-I*b)^(1/3)*gamma(2/3, (-I*b*d*x - I*b*c)*sqrt(d*x + c)))/(b^2*d^3)
\[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=\int \left (e + f x\right )^{2} \sin {\left (a + b c \sqrt {c + d x} + b d x \sqrt {c + d x} \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 694 vs. \(2 (290) = 580\).
Time = 0.52 (sec) , antiderivative size = 694, normalized size of antiderivative = 1.82 \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=\text {Too large to display} \]
-1/18*(3*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*(d*x + c)^ (3/2)*b) + (sqrt(3) - I)*gamma(2/3, -I*(d*x + c)^(3/2)*b))*cos(a) - ((I*sq rt(3) - 1)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3, - I*(d*x + c)^(3/2)*b))*sin(a))*e^2/(sqrt(d*x + c)*b) - 6*((d*x + c)^(3/2)*b )^(1/3)*(((sqrt(3) + I)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*ga mma(2/3, -I*(d*x + c)^(3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I*(d* x + c)^(3/2)*b) + (-I*sqrt(3) - 1)*gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(a ))*c*e*f/(sqrt(d*x + c)*b*d) + 3*((d*x + c)^(3/2)*b)^(1/3)*(((sqrt(3) + I) *gamma(2/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) - I)*gamma(2/3, -I*(d*x + c)^( 3/2)*b))*cos(a) - ((I*sqrt(3) - 1)*gamma(2/3, I*(d*x + c)^(3/2)*b) + (-I*s qrt(3) - 1)*gamma(2/3, -I*(d*x + c)^(3/2)*b))*sin(a))*c^2*f^2/(sqrt(d*x + c)*b*d^2) + 2*(12*((d*x + c)^(3/2)*b)^(1/3)*sqrt(d*x + c)*cos((d*x + c)^(3 /2)*b + a) + sqrt(d*x + c)*(((sqrt(3) - I)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) + I)*gamma(1/3, -I*(d*x + c)^(3/2)*b))*cos(a) + ((-I*sqrt(3) - 1)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (I*sqrt(3) - 1)*gamma(1/3, -I*(d*x + c)^(3/2)*b))*sin(a)))*e*f/(((d*x + c)^(3/2)*b)^(1/3)*b*d) - 2*(12*((d*x + c)^(3/2)*b)^(1/3)*sqrt(d*x + c)*cos((d*x + c)^(3/2)*b + a) + sqrt(d*x + c )*(((sqrt(3) - I)*gamma(1/3, I*(d*x + c)^(3/2)*b) + (sqrt(3) + I)*gamma(1/ 3, -I*(d*x + c)^(3/2)*b))*cos(a) + ((-I*sqrt(3) - 1)*gamma(1/3, I*(d*x + c )^(3/2)*b) + (I*sqrt(3) - 1)*gamma(1/3, -I*(d*x + c)^(3/2)*b))*sin(a)))...
\[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=\int { {\left (f x + e\right )}^{2} \sin \left ({\left (d x + c\right )}^{\frac {3}{2}} b + a\right ) \,d x } \]
Timed out. \[ \int (e+f x)^2 \sin \left (a+b (c+d x)^{3/2}\right ) \, dx=\int \sin \left (a+b\,{\left (c+d\,x\right )}^{3/2}\right )\,{\left (e+f\,x\right )}^2 \,d x \]