3.1.64 \(\int (a+b x^2)^{5/2} (c+d x^2) \, dx\) [64]

Optimal. Leaf size=149 \[ \frac {5 a^2 (8 b c-a d) x \sqrt {a+b x^2}}{128 b}+\frac {5 a (8 b c-a d) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {5 a^3 (8 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}} \]

[Out]

5/192*a*(-a*d+8*b*c)*x*(b*x^2+a)^(3/2)/b+1/48*(-a*d+8*b*c)*x*(b*x^2+a)^(5/2)/b+1/8*d*x*(b*x^2+a)^(7/2)/b+5/128
*a^3*(-a*d+8*b*c)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)+5/128*a^2*(-a*d+8*b*c)*x*(b*x^2+a)^(1/2)/b

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Rubi [A]
time = 0.03, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {396, 201, 223, 212} \begin {gather*} \frac {5 a^3 (8 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}+\frac {5 a^2 x \sqrt {a+b x^2} (8 b c-a d)}{128 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b c-a d)}{48 b}+\frac {5 a x \left (a+b x^2\right )^{3/2} (8 b c-a d)}{192 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)*(c + d*x^2),x]

[Out]

(5*a^2*(8*b*c - a*d)*x*Sqrt[a + b*x^2])/(128*b) + (5*a*(8*b*c - a*d)*x*(a + b*x^2)^(3/2))/(192*b) + ((8*b*c -
a*d)*x*(a + b*x^2)^(5/2))/(48*b) + (d*x*(a + b*x^2)^(7/2))/(8*b) + (5*a^3*(8*b*c - a*d)*ArcTanh[(Sqrt[b]*x)/Sq
rt[a + b*x^2]])/(128*b^(3/2))

Rule 201

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 212

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

Rule 223

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) \, dx &=\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {(-8 b c+a d) \int \left (a+b x^2\right )^{5/2} \, dx}{8 b}\\ &=\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {(5 a (8 b c-a d)) \int \left (a+b x^2\right )^{3/2} \, dx}{48 b}\\ &=\frac {5 a (8 b c-a d) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {\left (5 a^2 (8 b c-a d)\right ) \int \sqrt {a+b x^2} \, dx}{64 b}\\ &=\frac {5 a^2 (8 b c-a d) x \sqrt {a+b x^2}}{128 b}+\frac {5 a (8 b c-a d) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {\left (5 a^3 (8 b c-a d)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b}\\ &=\frac {5 a^2 (8 b c-a d) x \sqrt {a+b x^2}}{128 b}+\frac {5 a (8 b c-a d) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {\left (5 a^3 (8 b c-a d)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{128 b}\\ &=\frac {5 a^2 (8 b c-a d) x \sqrt {a+b x^2}}{128 b}+\frac {5 a (8 b c-a d) x \left (a+b x^2\right )^{3/2}}{192 b}+\frac {(8 b c-a d) x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {d x \left (a+b x^2\right )^{7/2}}{8 b}+\frac {5 a^3 (8 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 123, normalized size = 0.83 \begin {gather*} \frac {x \sqrt {a+b x^2} \left (264 a^2 b c+15 a^3 d+208 a b^2 c x^2+118 a^2 b d x^2+64 b^3 c x^4+136 a b^2 d x^4+48 b^3 d x^6\right )}{384 b}+\frac {5 a^3 (-8 b c+a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{128 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(264*a^2*b*c + 15*a^3*d + 208*a*b^2*c*x^2 + 118*a^2*b*d*x^2 + 64*b^3*c*x^4 + 136*a*b^2*d*x^
4 + 48*b^3*d*x^6))/(384*b) + (5*a^3*(-8*b*c + a*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(128*b^(3/2))

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Maple [A]
time = 0.05, size = 162, normalized size = 1.09

method	result	size
risch	\(\frac {x \left (48 b^{3} d \,x^{6}+136 a \,b^{2} d \,x^{4}+64 b^{3} c \,x^{4}+118 a^{2} b d \,x^{2}+208 a \,b^{2} c \,x^{2}+15 a^{3} d +264 a^{2} b c \right ) \sqrt {b \,x^{2}+a}}{384 b}-\frac {5 a^{4} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) d}{128 b^{\frac {3}{2}}}+\frac {5 a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c}{16 \sqrt {b}}\)	\(129\)
default	\(d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )+c \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

d*(1/8*x*(b*x^2+a)^(7/2)/b-1/8*a/b*(1/6*x*(b*x^2+a)^(5/2)+5/6*a*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^
(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))))))+c*(1/6*x*(b*x^2+a)^(5/2)+5/6*a*(1/4*x*(b*x^2+a)^(3/2)+3/
4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))))

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Maxima [A]
time = 0.30, size = 151, normalized size = 1.01 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} c x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a c x + \frac {5}{16} \, \sqrt {b x^{2} + a} a^{2} c x + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} d x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a d x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a^{3} d x}{128 \, b} + \frac {5 \, a^{3} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} - \frac {5 \, a^{4} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^2 + a)^(5/2)*c*x + 5/24*(b*x^2 + a)^(3/2)*a*c*x + 5/16*sqrt(b*x^2 + a)*a^2*c*x + 1/8*(b*x^2 + a)^(7/2
)*d*x/b - 1/48*(b*x^2 + a)^(5/2)*a*d*x/b - 5/192*(b*x^2 + a)^(3/2)*a^2*d*x/b - 5/128*sqrt(b*x^2 + a)*a^3*d*x/b
+ 5/16*a^3*c*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 5/128*a^4*d*arcsinh(b*x/sqrt(a*b))/b^(3/2)

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Fricas [A]
time = 0.51, size = 260, normalized size = 1.74 \begin {gather*} \left [-\frac {15 \, {\left (8 \, a^{3} b c - a^{4} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, b^{4} d x^{7} + 8 \, {\left (8 \, b^{4} c + 17 \, a b^{3} d\right )} x^{5} + 2 \, {\left (104 \, a b^{3} c + 59 \, a^{2} b^{2} d\right )} x^{3} + 3 \, {\left (88 \, a^{2} b^{2} c + 5 \, a^{3} b d\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{2}}, -\frac {15 \, {\left (8 \, a^{3} b c - a^{4} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, b^{4} d x^{7} + 8 \, {\left (8 \, b^{4} c + 17 \, a b^{3} d\right )} x^{5} + 2 \, {\left (104 \, a b^{3} c + 59 \, a^{2} b^{2} d\right )} x^{3} + 3 \, {\left (88 \, a^{2} b^{2} c + 5 \, a^{3} b d\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(15*(8*a^3*b*c - a^4*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(48*b^4*d*x^7 + 8*
(8*b^4*c + 17*a*b^3*d)*x^5 + 2*(104*a*b^3*c + 59*a^2*b^2*d)*x^3 + 3*(88*a^2*b^2*c + 5*a^3*b*d)*x)*sqrt(b*x^2 +
a))/b^2, -1/384*(15*(8*a^3*b*c - a^4*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (48*b^4*d*x^7 + 8*(8*b^
4*c + 17*a*b^3*d)*x^5 + 2*(104*a*b^3*c + 59*a^2*b^2*d)*x^3 + 3*(88*a^2*b^2*c + 5*a^3*b*d)*x)*sqrt(b*x^2 + a))/
b^2]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (134) = 268\).
time = 37.84, size = 316, normalized size = 2.12 \begin {gather*} \frac {5 a^{\frac {7}{2}} d x}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{\frac {5}{2}} c x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 a^{\frac {5}{2}} c x}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {133 a^{\frac {5}{2}} d x^{3}}{384 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 a^{\frac {3}{2}} b c x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {127 a^{\frac {3}{2}} b d x^{5}}{192 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 \sqrt {a} b^{2} c x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 \sqrt {a} b^{2} d x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 a^{4} d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {3}{2}}} + \frac {5 a^{3} c \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {b^{3} c x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {b^{3} d x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a**(7/2)*d*x/(128*b*sqrt(1 + b*x**2/a)) + a**(5/2)*c*x*sqrt(1 + b*x**2/a)/2 + 3*a**(5/2)*c*x/(16*sqrt(1 + b*
x**2/a)) + 133*a**(5/2)*d*x**3/(384*sqrt(1 + b*x**2/a)) + 35*a**(3/2)*b*c*x**3/(48*sqrt(1 + b*x**2/a)) + 127*a
**(3/2)*b*d*x**5/(192*sqrt(1 + b*x**2/a)) + 17*sqrt(a)*b**2*c*x**5/(24*sqrt(1 + b*x**2/a)) + 23*sqrt(a)*b**2*d
*x**7/(48*sqrt(1 + b*x**2/a)) - 5*a**4*d*asinh(sqrt(b)*x/sqrt(a))/(128*b**(3/2)) + 5*a**3*c*asinh(sqrt(b)*x/sq
rt(a))/(16*sqrt(b)) + b**3*c*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a)) + b**3*d*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]
time = 0.56, size = 135, normalized size = 0.91 \begin {gather*} \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} d x^{2} + \frac {8 \, b^{8} c + 17 \, a b^{7} d}{b^{6}}\right )} x^{2} + \frac {104 \, a b^{7} c + 59 \, a^{2} b^{6} d}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (88 \, a^{2} b^{6} c + 5 \, a^{3} b^{5} d\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {5 \, {\left (8 \, a^{3} b c - a^{4} d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*(4*(6*b^2*d*x^2 + (8*b^8*c + 17*a*b^7*d)/b^6)*x^2 + (104*a*b^7*c + 59*a^2*b^6*d)/b^6)*x^2 + 3*(88*a^2
*b^6*c + 5*a^3*b^5*d)/b^6)*sqrt(b*x^2 + a)*x - 5/128*(8*a^3*b*c - a^4*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))
)/b^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,x^2+a\right )}^{5/2}\,\left (d\,x^2+c\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)*(c + d*x^2),x)

[Out]

int((a + b*x^2)^(5/2)*(c + d*x^2), x)

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