\(\int (d x)^m (b x+c x^2)^{5/2} \, dx\) [118]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x) \left (b x+c x^2\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-\frac {5}{2}-m,\frac {9}{2},1+\frac {c x}{b}\right )}{7 c^3 x^2} \]

[Out]

2/7*b^2*(-c*x/b)^(-1/2-m)*(d*x)^m*(c*x+b)*(c*x^2+b*x)^(5/2)*hypergeom([7/2, -5/2-m],[9/2],1+c*x/b)/c^3/x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {688, 69, 67} \[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 b^2 (b+c x) \left (b x+c x^2\right )^{5/2} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m-\frac {5}{2},\frac {9}{2},\frac {c x}{b}+1\right )}{7 c^3 x^2} \]

[In]

Int[(d*x)^m*(b*x + c*x^2)^(5/2),x]

[Out]

(2*b^2*(-((c*x)/b))^(-1/2 - m)*(d*x)^m*(b + c*x)*(b*x + c*x^2)^(5/2)*Hypergeometric2F1[7/2, -5/2 - m, 9/2, 1 +
 (c*x)/b])/(7*c^3*x^2)

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
 &&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]

Rule 688

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*
(b + c*x)^p)), Int[x^(m + p)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, m}, x] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{-\frac {5}{2}-m} (d x)^m \left (b x+c x^2\right )^{5/2}\right ) \int x^{\frac {5}{2}+m} (b+c x)^{5/2} \, dx}{(b+c x)^{5/2}} \\ & = \frac {\left (b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m \left (b x+c x^2\right )^{5/2}\right ) \int \left (-\frac {c x}{b}\right )^{\frac {5}{2}+m} (b+c x)^{5/2} \, dx}{c^2 x^2 (b+c x)^{5/2}} \\ & = \frac {2 b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x) \left (b x+c x^2\right )^{5/2} \, _2F_1\left (\frac {7}{2},-\frac {5}{2}-m;\frac {9}{2};1+\frac {c x}{b}\right )}{7 c^3 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 b^2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x)^3 \sqrt {x (b+c x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-\frac {5}{2}-m,\frac {9}{2},1+\frac {c x}{b}\right )}{7 c^3} \]

[In]

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

[Out]

(2*b^2*(-((c*x)/b))^(-1/2 - m)*(d*x)^m*(b + c*x)^3*Sqrt[x*(b + c*x)]*Hypergeometric2F1[7/2, -5/2 - m, 9/2, 1 +
 (c*x)/b])/(7*c^3)

Maple [F]

\[\int \left (d x \right )^{m} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{2}} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + b^2*x^2)*sqrt(c*x^2 + b*x)*(d*x)^m, x)

Sympy [F]

\[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\int \left (d x\right )^{m} \left (x \left (b + c x\right )\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{2}} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {5}{2}} \left (d x\right )^{m} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (b x+c x^2\right )^{5/2} \, dx=\int {\left (c\,x^2+b\,x\right )}^{5/2}\,{\left (d\,x\right )}^m \,d x \]

[In]

int((b*x + c*x^2)^(5/2)*(d*x)^m,x)

[Out]

int((b*x + c*x^2)^(5/2)*(d*x)^m, x)