3.38 \(\int (c+d (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=23 \[ \frac{2 (d (a+b x)+c)^{5/2}}{5 b d} \]

[Out]

(2*(c + d*(a + b*x))^(5/2))/(5*b*d)

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Rubi [A]  time = 0.0109621, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {33, 32} \[ \frac{2 (d (a+b x)+c)^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*(a + b*x))^(3/2),x]

[Out]

(2*(c + d*(a + b*x))^(5/2))/(5*b*d)

Rule 33

Int[((a_.) + (b_.)*(u_))^(m_), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(a + b*x)^m, x], x, u], x]
/; FreeQ[{a, b, m}, x] && LinearQ[u, x] && NeQ[u, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int (c+d (a+b x))^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int (c+d x)^{3/2} \, dx,x,a+b x\right )}{b}\\ &=\frac{2 (c+d (a+b x))^{5/2}}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.0146494, size = 23, normalized size = 1. \[ \frac{2 (d (a+b x)+c)^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*(a + b*x))^(3/2),x]

[Out]

(2*(c + d*(a + b*x))^(5/2))/(5*b*d)

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Maple [A]  time = 0.001, size = 20, normalized size = 0.9 \begin{align*}{\frac{2}{5\,bd} \left ( bdx+ad+c \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02746, size = 26, normalized size = 1.13 \begin{align*} \frac{2 \,{\left ({\left (b x + a\right )} d + c\right )}^{\frac{5}{2}}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*((b*x + a)*d + c)^(5/2)/(b*d)

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Fricas [B]  time = 1.60548, size = 131, normalized size = 5.7 \begin{align*} \frac{2 \,{\left (b^{2} d^{2} x^{2} + a^{2} d^{2} + 2 \, a c d + c^{2} + 2 \,{\left (a b d^{2} + b c d\right )} x\right )} \sqrt{b d x + a d + c}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^2*d^2*x^2 + a^2*d^2 + 2*a*c*d + c^2 + 2*(a*b*d^2 + b*c*d)*x)*sqrt(b*d*x + a*d + c)/(b*d)

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Sympy [A]  time = 5.31987, size = 153, normalized size = 6.65 \begin{align*} \begin{cases} c^{\frac{3}{2}} x & \text{for}\: d = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \left (a d + c\right )^{\frac{3}{2}} & \text{for}\: b = 0 \\\frac{2 a^{2} d \sqrt{a d + b d x + c}}{5 b} + \frac{4 a d x \sqrt{a d + b d x + c}}{5} + \frac{4 a c \sqrt{a d + b d x + c}}{5 b} + \frac{2 b d x^{2} \sqrt{a d + b d x + c}}{5} + \frac{4 c x \sqrt{a d + b d x + c}}{5} + \frac{2 c^{2} \sqrt{a d + b d x + c}}{5 b d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**(3/2)*x, Eq(d, 0) & (Eq(b, 0) | Eq(d, 0))), (x*(a*d + c)**(3/2), Eq(b, 0)), (2*a**2*d*sqrt(a*d +
b*d*x + c)/(5*b) + 4*a*d*x*sqrt(a*d + b*d*x + c)/5 + 4*a*c*sqrt(a*d + b*d*x + c)/(5*b) + 2*b*d*x**2*sqrt(a*d
+ b*d*x + c)/5 + 4*c*x*sqrt(a*d + b*d*x + c)/5 + 2*c**2*sqrt(a*d + b*d*x + c)/(5*b*d), True))

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Giac [A]  time = 1.16723, size = 26, normalized size = 1.13 \begin{align*} \frac{2 \,{\left (b d x + a d + c\right )}^{\frac{5}{2}}}{5 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b*d*x + a*d + c)^(5/2)/(b*d)