\(\int (c x)^m (b x^n)^{5/2} \, dx\) [162]

Optimal result

Integrand size = 15, antiderivative size = 31 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 (c x)^{1+m} \left (b x^n\right )^{5/2}}{c (2+2 m+5 n)} \]

[Out]

2*(c*x)^(1+m)*(b*x^n)^(5/2)/c/(2+2*m+5*n)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {15, 20, 30} \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 b^2 x^{2 n+1} \sqrt {b x^n} (c x)^m}{2 m+5 n+2} \]

[In]

Int[(c*x)^m*(b*x^n)^(5/2),x]

[Out]

(2*b^2*x^(1 + 2*n)*(c*x)^m*Sqrt[b*x^n])/(2 + 2*m + 5*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 20

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {x (c x)^m \left (b x^n\right )^{5/2}}{1+m+\frac {5 n}{2}} \]

[In]

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

[Out]

(x*(c*x)^m*(b*x^n)^(5/2))/(1 + m + (5*n)/2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {2 x \left (c x \right )^{m} \left (b \,x^{n}\right )^{\frac {5}{2}}}{2+2 m +5 n}\)	\(26\)
risch	\(\frac {2 b^{3} x \,x^{m} c^{m} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i c x \right ) \pi m \left (\operatorname {csgn}\left (i c x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i c x \right )+\operatorname {csgn}\left (i c \right )\right )}{2}} x^{3 n}}{\left (2+2 m +5 n \right ) \sqrt {b \,x^{n}}}\)	\(75\)

[In]

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

[Out]

2*x/(2+2*m+5*n)*(c*x)^m*(b*x^n)^(5/2)

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).

Time = 138.73 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\begin {cases} \frac {2 x \left (b x^{n}\right )^{\frac {5}{2}} \left (c x\right )^{m}}{2 m + 5 n + 2} & \text {for}\: m \neq - \frac {5 n}{2} - 1 \\x \left (b x^{n}\right )^{\frac {5}{2}} \left (c x\right )^{- \frac {5 n}{2} - 1} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*x*(b*x**n)**(5/2)*(c*x)**m/(2*m + 5*n + 2), Ne(m, -5*n/2 - 1)), (x*(b*x**n)**(5/2)*(c*x)**(-5*n/2
- 1)*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 \, b^{\frac {5}{2}} c^{m} x x^{m} {\left (x^{n}\right )}^{\frac {5}{2}}}{2 \, m + 5 \, n + 2} \]

[In]

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

[Out]

2*b^(5/2)*c^m*x*x^m*(x^n)^(5/2)/(2*m + 5*n + 2)

Giac [F]

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

[In]

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

[Out]

integrate((b*x^n)^(5/2)*(c*x)^m, x)

Mupad [B] (verification not implemented)

Time = 5.57 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2\,b^2\,x^{2\,n+1}\,\sqrt {b\,x^n}\,{\left (c\,x\right )}^m}{2\,m+5\,n+2} \]

[In]

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

[Out]

(2*b^2*x^(2*n + 1)*(b*x^n)^(1/2)*(c*x)^m)/(2*m + 5*n + 2)