\(\int (d x)^m (c x^2)^{3/2} (a+b x)^n \, dx\) [982]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {c (d x)^{4+m} \sqrt {c x^2} (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \operatorname {Hypergeometric2F1}\left (4+m,-n,5+m,-\frac {b x}{a}\right )}{d^4 (4+m) x} \]

[Out]

c*(d*x)^(4+m)*(b*x+a)^n*hypergeom([-n, 4+m],[5+m],-b*x/a)*(c*x^2)^(1/2)/d^4/(4+m)/x/((1+b*x/a)^n)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 16, 68, 66} \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {c \sqrt {c x^2} (d x)^{m+4} (a+b x)^n \left (\frac {b x}{a}+1\right )^{-n} \operatorname {Hypergeometric2F1}\left (m+4,-n,m+5,-\frac {b x}{a}\right )}{d^4 (m+4) x} \]

[In]

Int[(d*x)^m*(c*x^2)^(3/2)*(a + b*x)^n,x]

[Out]

(c*(d*x)^(4 + m)*Sqrt[c*x^2]*(a + b*x)^n*Hypergeometric2F1[4 + m, -n, 5 + m, -((b*x)/a)])/(d^4*(4 + m)*x*(1 +
(b*x)/a)^n)

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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(
x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c \sqrt {c x^2}\right ) \int x^3 (d x)^m (a+b x)^n \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int (d x)^{3+m} (a+b x)^n \, dx}{d^3 x} \\ & = \frac {\left (c \sqrt {c x^2} (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n}\right ) \int (d x)^{3+m} \left (1+\frac {b x}{a}\right )^n \, dx}{d^3 x} \\ & = \frac {c (d x)^{4+m} \sqrt {c x^2} (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \, _2F_1\left (4+m,-n;5+m;-\frac {b x}{a}\right )}{d^4 (4+m) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\frac {x (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \operatorname {Hypergeometric2F1}\left (4+m,-n,5+m,-\frac {b x}{a}\right )}{4+m} \]

[In]

Integrate[(d*x)^m*(c*x^2)^(3/2)*(a + b*x)^n,x]

[Out]

(x*(d*x)^m*(c*x^2)^(3/2)*(a + b*x)^n*Hypergeometric2F1[4 + m, -n, 5 + m, -((b*x)/a)])/((4 + m)*(1 + (b*x)/a)^n
)

Maple [F]

\[\int \left (d x \right )^{m} \left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{n}d x\]

[In]

int((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^n,x)

[Out]

int((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^n,x)

Fricas [F]

\[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\int { \left (c x^{2}\right )^{\frac {3}{2}} {\left (b x + a\right )}^{n} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^n,x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2)*(b*x + a)^n*(d*x)^m*c*x^2, x)

Sympy [F(-1)]

Timed out. \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\text {Timed out} \]

[In]

integrate((d*x)**m*(c*x**2)**(3/2)*(b*x+a)**n,x)

[Out]

Timed out

Maxima [F]

\[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\int { \left (c x^{2}\right )^{\frac {3}{2}} {\left (b x + a\right )}^{n} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^n,x, algorithm="maxima")

[Out]

integrate((c*x^2)^(3/2)*(b*x + a)^n*(d*x)^m, x)

Giac [F]

\[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\int { \left (c x^{2}\right )^{\frac {3}{2}} {\left (b x + a\right )}^{n} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2)^(3/2)*(b*x+a)^n,x, algorithm="giac")

[Out]

integrate((c*x^2)^(3/2)*(b*x + a)^n*(d*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (c x^2\right )^{3/2} (a+b x)^n \, dx=\int {\left (d\,x\right )}^m\,{\left (c\,x^2\right )}^{3/2}\,{\left (a+b\,x\right )}^n \,d x \]

[In]

int((d*x)^m*(c*x^2)^(3/2)*(a + b*x)^n,x)

[Out]

int((d*x)^m*(c*x^2)^(3/2)*(a + b*x)^n, x)