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

Optimal. Leaf size=65 \[ \frac {a c^2 \sqrt {c x^2} (d x)^{m+6}}{d^6 (m+6) x}+\frac {b c^2 \sqrt {c x^2} (d x)^{m+7}}{d^7 (m+7) x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {15, 16, 43} \[ \frac {a c^2 \sqrt {c x^2} (d x)^{m+6}}{d^6 (m+6) x}+\frac {b c^2 \sqrt {c x^2} (d x)^{m+7}}{d^7 (m+7) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2*(d*x)^(6 + m)*Sqrt[c*x^2])/(d^6*(6 + m)*x) + (b*c^2*(d*x)^(7 + m)*Sqrt[c*x^2])/(d^7*(7 + 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 43

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

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Mathematica [A]  time = 0.03, size = 38, normalized size = 0.58 \[ \frac {x \left (c x^2\right )^{5/2} (d x)^m (a (m+7)+b (m+6) x)}{(m+6) (m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 58, normalized size = 0.89 \[ \frac {{\left ({\left (b c^{2} m + 6 \, b c^{2}\right )} x^{6} + {\left (a c^{2} m + 7 \, a c^{2}\right )} x^{5}\right )} \sqrt {c x^{2}} \left (d x\right )^{m}}{m^{2} + 13 \, m + 42} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*c^2*m + 6*b*c^2)*x^6 + (a*c^2*m + 7*a*c^2)*x^5)*sqrt(c*x^2)*(d*x)^m/(m^2 + 13*m + 42)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Undef/Unsigned Inf encountered in limit

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maple [A]  time = 0.00, size = 40, normalized size = 0.62 \[ \frac {\left (b m x +a m +6 b x +7 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}} x \left (d x \right )^{m}}{\left (m +7\right ) \left (m +6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.52, size = 39, normalized size = 0.60 \[ \frac {b c^{\frac {5}{2}} d^{m} x^{7} x^{m}}{m + 7} + \frac {a c^{\frac {5}{2}} d^{m} x^{6} x^{m}}{m + 6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 44, normalized size = 0.68 \[ \frac {c^2\,x^5\,{\left (d\,x\right )}^m\,\sqrt {c\,x^2}\,\left (7\,a+a\,m+6\,b\,x+b\,m\,x\right )}{m^2+13\,m+42} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*x^5*(d*x)^m*(c*x^2)^(1/2)*(7*a + a*m + 6*b*x + b*m*x))/(13*m + m^2 + 42)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {\int \frac {a \left (c x^{2}\right )^{\frac {5}{2}}}{x^{7}}\, dx + \int \frac {b \left (c x^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx}{d^{7}} & \text {for}\: m = -7 \\\frac {\int \frac {a \left (c x^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx + \int \frac {b \left (c x^{2}\right )^{\frac {5}{2}}}{x^{5}}\, dx}{d^{6}} & \text {for}\: m = -6 \\\frac {a c^{\frac {5}{2}} d^{m} m x x^{m} \left (x^{2}\right )^{\frac {5}{2}}}{m^{2} + 13 m + 42} + \frac {7 a c^{\frac {5}{2}} d^{m} x x^{m} \left (x^{2}\right )^{\frac {5}{2}}}{m^{2} + 13 m + 42} + \frac {b c^{\frac {5}{2}} d^{m} m x^{2} x^{m} \left (x^{2}\right )^{\frac {5}{2}}}{m^{2} + 13 m + 42} + \frac {6 b c^{\frac {5}{2}} d^{m} x^{2} x^{m} \left (x^{2}\right )^{\frac {5}{2}}}{m^{2} + 13 m + 42} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((Integral(a*(c*x**2)**(5/2)/x**7, x) + Integral(b*(c*x**2)**(5/2)/x**6, x))/d**7, Eq(m, -7)), ((Int
egral(a*(c*x**2)**(5/2)/x**6, x) + Integral(b*(c*x**2)**(5/2)/x**5, x))/d**6, Eq(m, -6)), (a*c**(5/2)*d**m*m*x
*x**m*(x**2)**(5/2)/(m**2 + 13*m + 42) + 7*a*c**(5/2)*d**m*x*x**m*(x**2)**(5/2)/(m**2 + 13*m + 42) + b*c**(5/2
)*d**m*m*x**2*x**m*(x**2)**(5/2)/(m**2 + 13*m + 42) + 6*b*c**(5/2)*d**m*x**2*x**m*(x**2)**(5/2)/(m**2 + 13*m +
42), True))

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