3.7.33 \(\int (a+b x)^{5/2} (c+d x)^{3/2} \, dx\)

Optimal. Leaf size=227 \[ -\frac {3 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{7/2}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}{128 b^2 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}{64 b^2 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}{80 b^2 d}+\frac {3 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b} \]

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Rubi [A]  time = 0.13, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {50, 63, 217, 206} \begin {gather*} \frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}{128 b^2 d^3}-\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}{64 b^2 d^2}-\frac {3 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{7/2}}+\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}{80 b^2 d}+\frac {3 (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)*(c + d*x)^(3/2),x]

[Out]

(3*(b*c - a*d)^4*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b^2*d^3) - ((b*c - a*d)^3*(a + b*x)^(3/2)*Sqrt[c + d*x])/(6
4*b^2*d^2) + ((b*c - a*d)^2*(a + b*x)^(5/2)*Sqrt[c + d*x])/(80*b^2*d) + (3*(b*c - a*d)*(a + b*x)^(7/2)*Sqrt[c
+ d*x])/(40*b^2) + ((a + b*x)^(7/2)*(c + d*x)^(3/2))/(5*b) - (3*(b*c - a*d)^5*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/
(Sqrt[b]*Sqrt[c + d*x])])/(128*b^(5/2)*d^(7/2))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

\begin {align*} \int (a+b x)^{5/2} (c+d x)^{3/2} \, dx &=\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}+\frac {(3 (b c-a d)) \int (a+b x)^{5/2} \sqrt {c+d x} \, dx}{10 b}\\ &=\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}+\frac {\left (3 (b c-a d)^2\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{80 b^2}\\ &=\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {(b c-a d)^3 \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{32 b^2 d}\\ &=-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}+\frac {\left (3 (b c-a d)^4\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{128 b^2 d^2}\\ &=\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^3}-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {\left (3 (b c-a d)^5\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{256 b^2 d^3}\\ &=\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^3}-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {\left (3 (b c-a d)^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{128 b^3 d^3}\\ &=\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^3}-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {\left (3 (b c-a d)^5\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{128 b^3 d^3}\\ &=\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^2 d^3}-\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{80 b^2 d}+\frac {3 (b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b^2}+\frac {(a+b x)^{7/2} (c+d x)^{3/2}}{5 b}-\frac {3 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 187, normalized size = 0.82 \begin {gather*} \frac {(a+b x)^{7/2} \sqrt {c+d x} \left (-\frac {15 (b c-a d)^{9/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{d^{7/2} (a+b x)^{7/2} \sqrt {\frac {b (c+d x)}{b c-a d}}}+\frac {15 (b c-a d)^4}{d^3 (a+b x)^3}+\frac {10 (a d-b c)^3}{d^2 (a+b x)^2}+\frac {8 (b c-a d)^2}{d (a+b x)}+48 (b c-a d)+128 b (c+d x)\right )}{640 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)*(c + d*x)^(3/2),x]

[Out]

((a + b*x)^(7/2)*Sqrt[c + d*x]*(48*(b*c - a*d) + (15*(b*c - a*d)^4)/(d^3*(a + b*x)^3) + (10*(-(b*c) + a*d)^3)/
(d^2*(a + b*x)^2) + (8*(b*c - a*d)^2)/(d*(a + b*x)) + 128*b*(c + d*x) - (15*(b*c - a*d)^(9/2)*ArcSinh[(Sqrt[d]
*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(d^(7/2)*(a + b*x)^(7/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(640*b^2)

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IntegrateAlgebraic [A]  time = 0.00, size = 197, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+b x} (b c-a d)^5 \left (-\frac {70 b^3 d (a+b x)}{c+d x}+\frac {128 b^2 d^2 (a+b x)^2}{(c+d x)^2}-\frac {15 d^4 (a+b x)^4}{(c+d x)^4}+\frac {70 b d^3 (a+b x)^3}{(c+d x)^3}+15 b^4\right )}{640 b^2 d^3 \sqrt {c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^5}-\frac {3 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{5/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(5/2)*(c + d*x)^(3/2),x]

[Out]

((b*c - a*d)^5*Sqrt[a + b*x]*(15*b^4 - (15*d^4*(a + b*x)^4)/(c + d*x)^4 + (70*b*d^3*(a + b*x)^3)/(c + d*x)^3 +
(128*b^2*d^2*(a + b*x)^2)/(c + d*x)^2 - (70*b^3*d*(a + b*x))/(c + d*x)))/(640*b^2*d^3*Sqrt[c + d*x]*(b - (d*(
a + b*x))/(c + d*x))^5) - (3*(b*c - a*d)^5*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(128*b^(5
/2)*d^(7/2))

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fricas [A]  time = 1.15, size = 702, normalized size = 3.09 \begin {gather*} \left [-\frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} + 70 \, a^{3} b^{2} c d^{4} - 15 \, a^{4} b d^{5} + 16 \, {\left (11 \, b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + 31 \, a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 23 \, a b^{4} c^{2} d^{3} - 233 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, b^{3} d^{4}}, \frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (128 \, b^{5} d^{5} x^{4} + 15 \, b^{5} c^{4} d - 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} + 70 \, a^{3} b^{2} c d^{4} - 15 \, a^{4} b d^{5} + 16 \, {\left (11 \, b^{5} c d^{4} + 21 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + 31 \, a^{2} b^{3} d^{5}\right )} x^{2} - 2 \, {\left (5 \, b^{5} c^{3} d^{2} - 23 \, a b^{4} c^{2} d^{3} - 233 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, b^{3} d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/2560*(15*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*sqr
t(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqr
t(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 70*a*b^4*c^3*d^2 + 128*a^2*b^3*c^2
*d^3 + 70*a^3*b^2*c*d^4 - 15*a^4*b*d^5 + 16*(11*b^5*c*d^4 + 21*a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 64*a*b^4*c*d^
4 + 31*a^2*b^3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 23*a*b^4*c^2*d^3 - 233*a^2*b^3*c*d^4 - 5*a^3*b^2*d^5)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(b^3*d^4), 1/1280*(15*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3
+ 5*a^4*b*c*d^4 - a^5*d^5)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/
(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(128*b^5*d^5*x^4 + 15*b^5*c^4*d - 70*a*b^4*c^3*d^2 + 128*
a^2*b^3*c^2*d^3 + 70*a^3*b^2*c*d^4 - 15*a^4*b*d^5 + 16*(11*b^5*c*d^4 + 21*a*b^4*d^5)*x^3 + 8*(b^5*c^2*d^3 + 64
*a*b^4*c*d^4 + 31*a^2*b^3*d^5)*x^2 - 2*(5*b^5*c^3*d^2 - 23*a*b^4*c^2*d^3 - 233*a^2*b^3*c*d^4 - 5*a^3*b^2*d^5)*
x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^4)]

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giac [B]  time = 3.46, size = 1740, normalized size = 7.67

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(3/2),x, algorithm="giac")

[Out]

1/1920*(240*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13
*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*
d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(
b*d)*b*d^2))*a*c*abs(b) - 1920*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d
- a*b*d)))/sqrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a))*a^3*c*abs(b)/b^2 + 10*(sqrt(b^2*c +
(b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) -
(5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15
*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^
2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqr
t(b*d)*b^2*d^3))*b*c*abs(b) + 30*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b
^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^
6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) +
3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x +
a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*a*d*abs(b) + 240*(sqrt(b^2*c + (b*x + a)*b*d -
a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 +
2*a*b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-s
qrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*a^2*d*abs(b)/b + (sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d)*(2*(4*(b*x + a)*(6*(b*x + a)*(8*(b*x + a)/b^4 + (b^20*c*d^7 - 41*a*b^19*d^8)/(b^23*d
^8)) - (7*b^21*c^2*d^6 + 26*a*b^20*c*d^7 - 513*a^2*b^19*d^8)/(b^23*d^8)) + 5*(7*b^22*c^3*d^5 + 19*a*b^21*c^2*d
^6 + 37*a^2*b^20*c*d^7 - 447*a^3*b^19*d^8)/(b^23*d^8))*(b*x + a) - 15*(7*b^23*c^4*d^4 + 12*a*b^22*c^3*d^5 + 18
*a^2*b^21*c^2*d^6 + 28*a^3*b^20*c*d^7 - 193*a^4*b^19*d^8)/(b^23*d^8))*sqrt(b*x + a) - 15*(7*b^5*c^5 + 5*a*b^4*
c^4*d + 6*a^2*b^3*c^3*d^2 + 10*a^3*b^2*c^2*d^3 + 35*a^4*b*c*d^4 - 63*a^5*d^5)*log(abs(-sqrt(b*d)*sqrt(b*x + a)
+ sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^3*d^4))*b*d*abs(b) + 1440*(sqrt(b^2*c + (b*x + a)*b*d -
a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sq
rt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a^2*c*abs(b)/b^2 + 480*(sqrt(b^2*
c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^
2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a^3*d*abs(b)/
b^3)/b

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maple [B]  time = 0.00, size = 853, normalized size = 3.76 \begin {gather*} \frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{5} d^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{256 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b^{2}}-\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{4} c d \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{256 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b}+\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} c^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{128 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}-\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b \,c^{3} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{128 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d}+\frac {15 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{2} c^{4} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{256 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d^{2}}-\frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} c^{5} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{256 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d^{3}}-\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{4} d}{128 b^{2}}+\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} c}{32 b}-\frac {9 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} c^{2}}{64 d}+\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, a b \,c^{3}}{32 d^{2}}-\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{2} c^{4}}{128 d^{3}}+\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a^{3}}{64 b}-\frac {3 \left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a^{2} c}{64 d}+\frac {3 \left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a b \,c^{2}}{64 d^{2}}-\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, b^{2} c^{3}}{64 d^{3}}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}} a^{2}}{16 d}-\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}} a b c}{8 d^{2}}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}} b^{2} c^{2}}{16 d^{3}}+\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}} a}{8 d}-\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}} b c}{8 d^{2}}+\frac {\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{2}}}{5 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(3/2),x)

[Out]

3/256*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a^5/b^2*d^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(
b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-15/256*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(
1/2)*a^4/b*c*d*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+15/128*((b*x+a)*(d*x+c)
)^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a^3*c^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a
*d+b*c)*x)^(1/2))-15/128*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a^2*b*c^3/d*ln((b*d*x
+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+15/256*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b
*x+a)^(1/2)/(b*d)^(1/2)*a*b^2*c^4/d^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-
3/256*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*b^3*c^5/d^3*ln((b*d*x+1/2*a*d+1/2*b*c)/(
b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-3/128*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^4/b^2*d+3/32*(d*x+c)^(1/2)*(b*
x+a)^(1/2)*a^3/b*c-9/64*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2*c^2/d+3/32*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*b*c^3/d^2-3/1
28*(d*x+c)^(1/2)*(b*x+a)^(1/2)*b^2*c^4/d^3+1/64*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a^3/b-3/64*(d*x+c)^(3/2)*(b*x+a)^(
1/2)*a^2*c/d+3/64*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a*b*c^2/d^2-1/64*(d*x+c)^(3/2)*(b*x+a)^(1/2)*b^2*c^3/d^3+1/16*(b
*x+a)^(1/2)*(d*x+c)^(5/2)*a^2/d-1/8*(b*x+a)^(1/2)*(d*x+c)^(5/2)*a*b*c/d^2+1/16*(b*x+a)^(1/2)*(d*x+c)^(5/2)*b^2
*c^2/d^3+1/8*(b*x+a)^(3/2)*(d*x+c)^(5/2)*a/d-1/8*(b*x+a)^(3/2)*(d*x+c)^(5/2)*b*c/d^2+1/5*(b*x+a)^(5/2)*(d*x+c)
^(5/2)/d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)*(c + d*x)^(3/2),x)

[Out]

int((a + b*x)^(5/2)*(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(3/2),x)

[Out]

Integral((a + b*x)**(5/2)*(c + d*x)**(3/2), x)

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