\(\int (d x)^m (c x^2)^{5/2} (a+b x)^2 \, dx\) [975]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F(-2)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\frac {a^2 c^2 (d x)^{6+m} \sqrt {c x^2}}{d^6 (6+m) x}+\frac {2 a b c^2 (d x)^{7+m} \sqrt {c x^2}}{d^7 (7+m) x}+\frac {b^2 c^2 (d x)^{8+m} \sqrt {c x^2}}{d^8 (8+m) x} \]

[Out]

a^2*c^2*(d*x)^(6+m)*(c*x^2)^(1/2)/d^6/(6+m)/x+2*a*b*c^2*(d*x)^(7+m)*(c*x^2)^(1/2)/d^7/(7+m)/x+b^2*c^2*(d*x)^(8
+m)*(c*x^2)^(1/2)/d^8/(8+m)/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 16, 45} \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\frac {a^2 c^2 \sqrt {c x^2} (d x)^{m+6}}{d^6 (m+6) x}+\frac {2 a b c^2 \sqrt {c x^2} (d x)^{m+7}}{d^7 (m+7) x}+\frac {b^2 c^2 \sqrt {c x^2} (d x)^{m+8}}{d^8 (m+8) x} \]

[In]

Int[(d*x)^m*(c*x^2)^(5/2)*(a + b*x)^2,x]

[Out]

(a^2*c^2*(d*x)^(6 + m)*Sqrt[c*x^2])/(d^6*(6 + m)*x) + (2*a*b*c^2*(d*x)^(7 + m)*Sqrt[c*x^2])/(d^7*(7 + m)*x) +
(b^2*c^2*(d*x)^(8 + m)*Sqrt[c*x^2])/(d^8*(8 + m)*x)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c^2 \sqrt {c x^2}\right ) \int x^5 (d x)^m (a+b x)^2 \, dx}{x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int (d x)^{5+m} (a+b x)^2 \, dx}{d^5 x} \\ & = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a^2 (d x)^{5+m}+\frac {2 a b (d x)^{6+m}}{d}+\frac {b^2 (d x)^{7+m}}{d^2}\right ) \, dx}{d^5 x} \\ & = \frac {a^2 c^2 (d x)^{6+m} \sqrt {c x^2}}{d^6 (6+m) x}+\frac {2 a b c^2 (d x)^{7+m} \sqrt {c x^2}}{d^7 (7+m) x}+\frac {b^2 c^2 (d x)^{8+m} \sqrt {c x^2}}{d^8 (8+m) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.47 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=x (d x)^m \left (c x^2\right )^{5/2} \left (\frac {a^2}{6+m}+\frac {2 a b x}{7+m}+\frac {b^2 x^2}{8+m}\right ) \]

[In]

Integrate[(d*x)^m*(c*x^2)^(5/2)*(a + b*x)^2,x]

[Out]

x*(d*x)^m*(c*x^2)^(5/2)*(a^2/(6 + m) + (2*a*b*x)/(7 + m) + (b^2*x^2)/(8 + m))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.92

method result size
gosper \(\frac {x \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +13 m \,x^{2} b^{2}+a^{2} m^{2}+28 a b m x +42 b^{2} x^{2}+15 a^{2} m +96 a b x +56 a^{2}\right ) \left (d x \right )^{m} \left (c \,x^{2}\right )^{\frac {5}{2}}}{\left (8+m \right ) \left (7+m \right ) \left (6+m \right )}\) \(95\)
risch \(\frac {c^{2} x^{5} \sqrt {c \,x^{2}}\, \left (b^{2} m^{2} x^{2}+2 a b \,m^{2} x +13 m \,x^{2} b^{2}+a^{2} m^{2}+28 a b m x +42 b^{2} x^{2}+15 a^{2} m +96 a b x +56 a^{2}\right ) \left (d x \right )^{m}}{\left (8+m \right ) \left (7+m \right ) \left (6+m \right )}\) \(100\)

[In]

int((d*x)^m*(c*x^2)^(5/2)*(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

x*(b^2*m^2*x^2+2*a*b*m^2*x+13*b^2*m*x^2+a^2*m^2+28*a*b*m*x+42*b^2*x^2+15*a^2*m+96*a*b*x+56*a^2)*(d*x)^m*(c*x^2
)^(5/2)/(8+m)/(7+m)/(6+m)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.19 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\frac {{\left ({\left (b^{2} c^{2} m^{2} + 13 \, b^{2} c^{2} m + 42 \, b^{2} c^{2}\right )} x^{7} + 2 \, {\left (a b c^{2} m^{2} + 14 \, a b c^{2} m + 48 \, a b c^{2}\right )} x^{6} + {\left (a^{2} c^{2} m^{2} + 15 \, a^{2} c^{2} m + 56 \, a^{2} c^{2}\right )} x^{5}\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{3} + 21 \, m^{2} + 146 \, m + 336} \]

[In]

integrate((d*x)^m*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="fricas")

[Out]

((b^2*c^2*m^2 + 13*b^2*c^2*m + 42*b^2*c^2)*x^7 + 2*(a*b*c^2*m^2 + 14*a*b*c^2*m + 48*a*b*c^2)*x^6 + (a^2*c^2*m^
2 + 15*a^2*c^2*m + 56*a^2*c^2)*x^5)*sqrt(c*x^2)*(d*x)^m/(m^3 + 21*m^2 + 146*m + 336)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (92) = 184\).

Time = 12.09 (sec) , antiderivative size = 495, normalized size of antiderivative = 4.81 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\begin {cases} \frac {- \frac {a^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{2 x^{7}} - \frac {2 a b \left (c x^{2}\right )^{\frac {5}{2}}}{x^{6}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {5}{2}} \log {\left (x \right )}}{x^{5}}}{d^{8}} & \text {for}\: m = -8 \\\frac {- \frac {a^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{x^{6}} + \frac {2 a b \left (c x^{2}\right )^{\frac {5}{2}} \log {\left (x \right )}}{x^{5}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{x^{4}}}{d^{7}} & \text {for}\: m = -7 \\\frac {\frac {a^{2} \left (c x^{2}\right )^{\frac {5}{2}} \log {\left (x \right )}}{x^{5}} + \frac {2 a b \left (c x^{2}\right )^{\frac {5}{2}}}{x^{4}} + \frac {b^{2} \left (c x^{2}\right )^{\frac {5}{2}}}{2 x^{3}}}{d^{6}} & \text {for}\: m = -6 \\\frac {a^{2} m^{2} x \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {15 a^{2} m x \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {56 a^{2} x \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {2 a b m^{2} x^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {28 a b m x^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {96 a b x^{2} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {b^{2} m^{2} x^{3} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {13 b^{2} m x^{3} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} + \frac {42 b^{2} x^{3} \left (c x^{2}\right )^{\frac {5}{2}} \left (d x\right )^{m}}{m^{3} + 21 m^{2} + 146 m + 336} & \text {otherwise} \end {cases} \]

[In]

integrate((d*x)**m*(c*x**2)**(5/2)*(b*x+a)**2,x)

[Out]

Piecewise(((-a**2*(c*x**2)**(5/2)/(2*x**7) - 2*a*b*(c*x**2)**(5/2)/x**6 + b**2*(c*x**2)**(5/2)*log(x)/x**5)/d*
*8, Eq(m, -8)), ((-a**2*(c*x**2)**(5/2)/x**6 + 2*a*b*(c*x**2)**(5/2)*log(x)/x**5 + b**2*(c*x**2)**(5/2)/x**4)/
d**7, Eq(m, -7)), ((a**2*(c*x**2)**(5/2)*log(x)/x**5 + 2*a*b*(c*x**2)**(5/2)/x**4 + b**2*(c*x**2)**(5/2)/(2*x*
*3))/d**6, Eq(m, -6)), (a**2*m**2*x*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 15*a**2*m*x*(c*x
**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 56*a**2*x*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146
*m + 336) + 2*a*b*m**2*x**2*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 28*a*b*m*x**2*(c*x**2)**
(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 96*a*b*x**2*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m +
 336) + b**2*m**2*x**3*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 13*b**2*m*x**3*(c*x**2)**(5/2
)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 336) + 42*b**2*x**3*(c*x**2)**(5/2)*(d*x)**m/(m**3 + 21*m**2 + 146*m + 33
6), True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.62 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\frac {b^{2} c^{\frac {5}{2}} d^{m} x^{8} x^{m}}{m + 8} + \frac {2 \, a b c^{\frac {5}{2}} d^{m} x^{7} x^{m}}{m + 7} + \frac {a^{2} c^{\frac {5}{2}} d^{m} x^{6} x^{m}}{m + 6} \]

[In]

integrate((d*x)^m*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="maxima")

[Out]

b^2*c^(5/2)*d^m*x^8*x^m/(m + 8) + 2*a*b*c^(5/2)*d^m*x^7*x^m/(m + 7) + a^2*c^(5/2)*d^m*x^6*x^m/(m + 6)

Giac [F(-2)]

Exception generated. \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((d*x)^m*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.23 \[ \int (d x)^m \left (c x^2\right )^{5/2} (a+b x)^2 \, dx={\left (d\,x\right )}^m\,\left (\frac {a^2\,c^2\,x^5\,\sqrt {c\,x^2}\,\left (m^2+15\,m+56\right )}{m^3+21\,m^2+146\,m+336}+\frac {b^2\,c^2\,x^7\,\sqrt {c\,x^2}\,\left (m^2+13\,m+42\right )}{m^3+21\,m^2+146\,m+336}+\frac {2\,a\,b\,c^2\,x^6\,\sqrt {c\,x^2}\,\left (m^2+14\,m+48\right )}{m^3+21\,m^2+146\,m+336}\right ) \]

[In]

int((d*x)^m*(c*x^2)^(5/2)*(a + b*x)^2,x)

[Out]

(d*x)^m*((a^2*c^2*x^5*(c*x^2)^(1/2)*(15*m + m^2 + 56))/(146*m + 21*m^2 + m^3 + 336) + (b^2*c^2*x^7*(c*x^2)^(1/
2)*(13*m + m^2 + 42))/(146*m + 21*m^2 + m^3 + 336) + (2*a*b*c^2*x^6*(c*x^2)^(1/2)*(14*m + m^2 + 48))/(146*m +
21*m^2 + m^3 + 336))