3.7.59 \(\int (a+b x)^3 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx\)

Optimal. Leaf size=167 \[ \frac {7 a^2 b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{5 c}-\frac {7 a^5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2-\frac {a^2 c}{b^2}}}\right )}{8 b^2}+\frac {7}{8} a^3 x \sqrt {c x^2-\frac {a^2 c}{b^2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {671, 641, 195, 217, 206} \begin {gather*} \frac {7}{8} a^3 x \sqrt {c x^2-\frac {a^2 c}{b^2}}+\frac {7 a^2 b \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (c x^2-\frac {a^2 c}{b^2}\right )^{3/2}}{5 c}-\frac {7 a^5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2-\frac {a^2 c}{b^2}}}\right )}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*Sqrt[-((a^2*c)/b^2) + c*x^2],x]

[Out]

(7*a^3*x*Sqrt[-((a^2*c)/b^2) + c*x^2])/8 + (7*a^2*b*(-((a^2*c)/b^2) + c*x^2)^(3/2))/(12*c) + (7*a*b*(a + b*x)*
(-((a^2*c)/b^2) + c*x^2)^(3/2))/(20*c) + (b*(a + b*x)^2*(-((a^2*c)/b^2) + c*x^2)^(3/2))/(5*c) - (7*a^5*Sqrt[c]
*ArcTanh[(Sqrt[c]*x)/Sqrt[-((a^2*c)/b^2) + c*x^2]])/(8*b^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int (a+b x)^3 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx &=\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}+\frac {1}{5} (7 a) \int (a+b x)^2 \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx\\ &=\frac {7 a b (a+b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}+\frac {1}{4} \left (7 a^2\right ) \int (a+b x) \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx\\ &=\frac {7 a^2 b \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}+\frac {1}{4} \left (7 a^3\right ) \int \sqrt {-\frac {a^2 c}{b^2}+c x^2} \, dx\\ &=\frac {7}{8} a^3 x \sqrt {-\frac {a^2 c}{b^2}+c x^2}+\frac {7 a^2 b \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}-\frac {\left (7 a^5 c\right ) \int \frac {1}{\sqrt {-\frac {a^2 c}{b^2}+c x^2}} \, dx}{8 b^2}\\ &=\frac {7}{8} a^3 x \sqrt {-\frac {a^2 c}{b^2}+c x^2}+\frac {7 a^2 b \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}-\frac {\left (7 a^5 c\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {-\frac {a^2 c}{b^2}+c x^2}}\right )}{8 b^2}\\ &=\frac {7}{8} a^3 x \sqrt {-\frac {a^2 c}{b^2}+c x^2}+\frac {7 a^2 b \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{12 c}+\frac {7 a b (a+b x) \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{20 c}+\frac {b (a+b x)^2 \left (-\frac {a^2 c}{b^2}+c x^2\right )^{3/2}}{5 c}-\frac {7 a^5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {-\frac {a^2 c}{b^2}+c x^2}}\right )}{8 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 114, normalized size = 0.68 \begin {gather*} \frac {\sqrt {c \left (x^2-\frac {a^2}{b^2}\right )} \left (105 a^4 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-136 a^4+15 a^3 b x+112 a^2 b^2 x^2+90 a b^3 x^3+24 b^4 x^4\right )\right )}{120 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*Sqrt[-((a^2*c)/b^2) + c*x^2],x]

[Out]

(Sqrt[c*(-(a^2/b^2) + x^2)]*(Sqrt[1 - (b^2*x^2)/a^2]*(-136*a^4 + 15*a^3*b*x + 112*a^2*b^2*x^2 + 90*a*b^3*x^3 +
24*b^4*x^4) + 105*a^4*ArcSin[(b*x)/a]))/(120*b*Sqrt[1 - (b^2*x^2)/a^2])

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IntegrateAlgebraic [A]  time = 0.40, size = 112, normalized size = 0.67 \begin {gather*} \frac {7 a^5 \sqrt {c} \log \left (\sqrt {c x^2-\frac {a^2 c}{b^2}}-\sqrt {c} x\right )}{8 b^2}+\frac {\left (-136 a^4+15 a^3 b x+112 a^2 b^2 x^2+90 a b^3 x^3+24 b^4 x^4\right ) \sqrt {c x^2-\frac {a^2 c}{b^2}}}{120 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^3*Sqrt[-((a^2*c)/b^2) + c*x^2],x]

[Out]

(Sqrt[-((a^2*c)/b^2) + c*x^2]*(-136*a^4 + 15*a^3*b*x + 112*a^2*b^2*x^2 + 90*a*b^3*x^3 + 24*b^4*x^4))/(120*b) +
(7*a^5*Sqrt[c]*Log[-(Sqrt[c]*x) + Sqrt[-((a^2*c)/b^2) + c*x^2]])/(8*b^2)

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fricas [A]  time = 0.45, size = 260, normalized size = 1.56 \begin {gather*} \left [\frac {105 \, a^{5} \sqrt {c} \log \left (2 \, b^{2} c x^{2} - 2 \, b^{2} \sqrt {c} x \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}} - a^{2} c\right ) + 2 \, {\left (24 \, b^{5} x^{4} + 90 \, a b^{4} x^{3} + 112 \, a^{2} b^{3} x^{2} + 15 \, a^{3} b^{2} x - 136 \, a^{4} b\right )} \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{240 \, b^{2}}, \frac {105 \, a^{5} \sqrt {-c} \arctan \left (\frac {b^{2} \sqrt {-c} x \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (24 \, b^{5} x^{4} + 90 \, a b^{4} x^{3} + 112 \, a^{2} b^{3} x^{2} + 15 \, a^{3} b^{2} x - 136 \, a^{4} b\right )} \sqrt {\frac {b^{2} c x^{2} - a^{2} c}{b^{2}}}}{120 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/240*(105*a^5*sqrt(c)*log(2*b^2*c*x^2 - 2*b^2*sqrt(c)*x*sqrt((b^2*c*x^2 - a^2*c)/b^2) - a^2*c) + 2*(24*b^5*x
^4 + 90*a*b^4*x^3 + 112*a^2*b^3*x^2 + 15*a^3*b^2*x - 136*a^4*b)*sqrt((b^2*c*x^2 - a^2*c)/b^2))/b^2, 1/120*(105
*a^5*sqrt(-c)*arctan(b^2*sqrt(-c)*x*sqrt((b^2*c*x^2 - a^2*c)/b^2)/(b^2*c*x^2 - a^2*c)) + (24*b^5*x^4 + 90*a*b^
4*x^3 + 112*a^2*b^3*x^2 + 15*a^3*b^2*x - 136*a^4*b)*sqrt((b^2*c*x^2 - a^2*c)/b^2))/b^2]

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giac [A]  time = 0.23, size = 113, normalized size = 0.68 \begin {gather*} \frac {{\left (\frac {105 \, a^{5} \sqrt {c} \log \left ({\left | -\sqrt {b^{2} c} x + \sqrt {b^{2} c x^{2} - a^{2} c} \right |}\right )}{{\left | b \right |}} - \sqrt {b^{2} c x^{2} - a^{2} c} {\left (\frac {136 \, a^{4}}{b} - {\left (15 \, a^{3} + 2 \, {\left (56 \, a^{2} b + 3 \, {\left (4 \, b^{3} x + 15 \, a b^{2}\right )} x\right )} x\right )} x\right )}\right )} {\left | b \right |}}{120 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="giac")

[Out]

1/120*(105*a^5*sqrt(c)*log(abs(-sqrt(b^2*c)*x + sqrt(b^2*c*x^2 - a^2*c)))/abs(b) - sqrt(b^2*c*x^2 - a^2*c)*(13
6*a^4/b - (15*a^3 + 2*(56*a^2*b + 3*(4*b^3*x + 15*a*b^2)*x)*x)*x))*abs(b)/b^2

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maple [A]  time = 0.05, size = 169, normalized size = 1.01 \begin {gather*} -\frac {7 a^{5} \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}-\frac {a^{2} c}{b^{2}}}\right )}{8 b^{2}}+\frac {7 \sqrt {c \,x^{2}-\frac {a^{2} c}{b^{2}}}\, a^{3} x}{8}+\frac {\left (c \,x^{2}-\frac {a^{2} c}{b^{2}}\right )^{\frac {3}{2}} b^{3} x^{2}}{5 c}+\frac {3 \left (c \,x^{2}-\frac {a^{2} c}{b^{2}}\right )^{\frac {3}{2}} a \,b^{2} x}{4 c}+\frac {2 \left (c \,x^{2}-\frac {a^{2} c}{b^{2}}\right )^{\frac {3}{2}} a^{2} b}{15 c}+\frac {\left (\frac {\left (b^{2} x^{2}-a^{2}\right ) c}{b^{2}}\right )^{\frac {3}{2}} a^{2} b}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(c*x^2-a^2/b^2*c)^(1/2),x)

[Out]

1/5*b^3*x^2*(c*x^2-a^2/b^2*c)^(3/2)/c+2/15*a^2*b*(c*x^2-a^2/b^2*c)^(3/2)/c+3/4*a*b^2*x*(c*x^2-a^2/b^2*c)^(3/2)
/c+7/8*a^3*x*(c*x^2-a^2/b^2*c)^(1/2)-7/8*a^5/b^2*c^(1/2)*ln(c^(1/2)*x+(c*x^2-a^2/b^2*c)^(1/2))+a^2*b/c*((b^2*x
^2-a^2)/b^2*c)^(3/2)

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maxima [A]  time = 1.35, size = 144, normalized size = 0.86 \begin {gather*} \frac {{\left (c x^{2} - \frac {a^{2} c}{b^{2}}\right )}^{\frac {3}{2}} b^{3} x^{2}}{5 \, c} + \frac {7}{8} \, \sqrt {c x^{2} - \frac {a^{2} c}{b^{2}}} a^{3} x + \frac {3 \, {\left (c x^{2} - \frac {a^{2} c}{b^{2}}\right )}^{\frac {3}{2}} a b^{2} x}{4 \, c} - \frac {7 \, a^{5} \sqrt {c} \log \left (2 \, c x + 2 \, \sqrt {c x^{2} - \frac {a^{2} c}{b^{2}}} \sqrt {c}\right )}{8 \, b^{2}} + \frac {17 \, {\left (c x^{2} - \frac {a^{2} c}{b^{2}}\right )}^{\frac {3}{2}} a^{2} b}{15 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(-a^2*c/b^2+c*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/5*(c*x^2 - a^2*c/b^2)^(3/2)*b^3*x^2/c + 7/8*sqrt(c*x^2 - a^2*c/b^2)*a^3*x + 3/4*(c*x^2 - a^2*c/b^2)^(3/2)*a*
b^2*x/c - 7/8*a^5*sqrt(c)*log(2*c*x + 2*sqrt(c*x^2 - a^2*c/b^2)*sqrt(c))/b^2 + 17/15*(c*x^2 - a^2*c/b^2)^(3/2)
*a^2*b/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {c\,x^2-\frac {a^2\,c}{b^2}}\,{\left (a+b\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2 - (a^2*c)/b^2)^(1/2)*(a + b*x)^3,x)

[Out]

int((c*x^2 - (a^2*c)/b^2)^(1/2)*(a + b*x)^3, x)

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sympy [C]  time = 7.82, size = 491, normalized size = 2.94 \begin {gather*} - \frac {2 a^{4} \sqrt {- \frac {a^{2} c}{b^{2}} + c x^{2}}}{15 b} + a^{3} \left (\begin {cases} - \frac {a^{2} \sqrt {c} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{2 b^{2}} - \frac {a \sqrt {c} x}{2 b \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {b \sqrt {c} x^{3}}{2 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {i a^{2} \sqrt {c} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{2 b^{2}} + \frac {i a \sqrt {c} x \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}}{2 b} & \text {otherwise} \end {cases}\right ) - \frac {a^{2} b x^{2} \sqrt {- \frac {a^{2} c}{b^{2}} + c x^{2}}}{15} + 3 a^{2} b \left (\begin {cases} 0 & \text {for}\: c = 0 \\\frac {\left (- \frac {a^{2} c}{b^{2}} + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + 3 a b^{2} \left (\begin {cases} - \frac {a^{4} \sqrt {c} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{8 b^{4}} + \frac {a^{3} \sqrt {c} x}{8 b^{3} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {3 a \sqrt {c} x^{3}}{8 b \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {b \sqrt {c} x^{5}}{4 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {i a^{4} \sqrt {c} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{8 b^{4}} - \frac {i a^{3} \sqrt {c} x}{8 b^{3} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {3 i a \sqrt {c} x^{3}}{8 b \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} - \frac {i b \sqrt {c} x^{5}}{4 a \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} & \text {otherwise} \end {cases}\right ) + \frac {b^{3} x^{4} \sqrt {- \frac {a^{2} c}{b^{2}} + c x^{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(-a**2*c/b**2+c*x**2)**(1/2),x)

[Out]

-2*a**4*sqrt(-a**2*c/b**2 + c*x**2)/(15*b) + a**3*Piecewise((-a**2*sqrt(c)*acosh(b*x/a)/(2*b**2) - a*sqrt(c)*x
/(2*b*sqrt(-1 + b**2*x**2/a**2)) + b*sqrt(c)*x**3/(2*a*sqrt(-1 + b**2*x**2/a**2)), Abs(b**2*x**2/a**2) > 1), (
I*a**2*sqrt(c)*asin(b*x/a)/(2*b**2) + I*a*sqrt(c)*x*sqrt(1 - b**2*x**2/a**2)/(2*b), True)) - a**2*b*x**2*sqrt(
-a**2*c/b**2 + c*x**2)/15 + 3*a**2*b*Piecewise((0, Eq(c, 0)), ((-a**2*c/b**2 + c*x**2)**(3/2)/(3*c), True)) +
3*a*b**2*Piecewise((-a**4*sqrt(c)*acosh(b*x/a)/(8*b**4) + a**3*sqrt(c)*x/(8*b**3*sqrt(-1 + b**2*x**2/a**2)) -
3*a*sqrt(c)*x**3/(8*b*sqrt(-1 + b**2*x**2/a**2)) + b*sqrt(c)*x**5/(4*a*sqrt(-1 + b**2*x**2/a**2)), Abs(b**2*x*
*2/a**2) > 1), (I*a**4*sqrt(c)*asin(b*x/a)/(8*b**4) - I*a**3*sqrt(c)*x/(8*b**3*sqrt(1 - b**2*x**2/a**2)) + 3*I
*a*sqrt(c)*x**3/(8*b*sqrt(1 - b**2*x**2/a**2)) - I*b*sqrt(c)*x**5/(4*a*sqrt(1 - b**2*x**2/a**2)), True)) + b**
3*x**4*sqrt(-a**2*c/b**2 + c*x**2)/5

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