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

Optimal. Leaf size=186 \[ -\frac {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{7/2} d^{3/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 b^3 d}+\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2}{32 b^3}+\frac {5 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b} \]

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Rubi [A]  time = 0.09, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{7/2} d^{3/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 b^3 d}+\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2}{32 b^3}+\frac {5 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(c + d*x)^(5/2),x]

[Out]

(5*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3*d) + (5*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(32
*b^3) + (5*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*b^2) + ((a + b*x)^(3/2)*(c + d*x)^(5/2))/(4*b) - (
5*(b*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(3/2))

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 \sqrt {a+b x} (c+d x)^{5/2} \, dx &=\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}+\frac {(5 (b c-a d)) \int \sqrt {a+b x} (c+d x)^{3/2} \, dx}{8 b}\\ &=\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}+\frac {\left (5 (b c-a d)^2\right ) \int \sqrt {a+b x} \sqrt {c+d x} \, dx}{16 b^2}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^3}+\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}+\frac {\left (5 (b c-a d)^3\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 b^3}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b^3 d}+\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^3}+\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}-\frac {\left (5 (b c-a d)^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b^3 d}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b^3 d}+\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^3}+\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}-\frac {\left (5 (b c-a d)^4\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 )}{64 b^4 d}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b^3 d}+\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^3}+\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}-\frac {\left (5 (b c-a d)^4\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 )}{64 b^4 d}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b^3 d}+\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^3}+\frac {5 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{5/2}}{4 b}-\frac {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{7/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 191, normalized size = 1.03 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (15 a^3 d^3-5 a^2 b d^2 (11 c+2 d x)+a b^2 d \left (73 c^2+36 c d x+8 d^2 x^2\right )+b^3 \left (15 c^3+118 c^2 d x+136 c d^2 x^2+48 d^3 x^3\right )\right )-15 (b c-a d)^{9/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{192 b^4 d^{3/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(15*a^3*d^3 - 5*a^2*b*d^2*(11*c + 2*d*x) + a*b^2*d*(73*c^2 + 36*c*d*x + 8*d
^2*x^2) + b^3*(15*c^3 + 118*c^2*d*x + 136*c*d^2*x^2 + 48*d^3*x^3)) - 15*(b*c - a*d)^(9/2)*Sqrt[(b*(c + d*x))/(
b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(192*b^4*d^(3/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.00, size = 175, normalized size = 0.94 \begin {gather*} \frac {\sqrt {a+b x} (b c-a d)^4 \left (\frac {73 b^2 d (a+b x)}{c+d x}+\frac {15 d^3 (a+b x)^3}{(c+d x)^3}-\frac {55 b d^2 (a+b x)^2}{(c+d x)^2}+15 b^3\right )}{192 b^3 d \sqrt {c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )^4}-\frac {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{7/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^4*Sqrt[a + b*x]*(15*b^3 + (15*d^3*(a + b*x)^3)/(c + d*x)^3 - (55*b*d^2*(a + b*x)^2)/(c + d*x)^2 +
(73*b^2*d*(a + b*x))/(c + d*x)))/(192*b^3*d*Sqrt[c + d*x]*(b - (d*(a + b*x))/(c + d*x))^4) - (5*(b*c - a*d)^4
*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(3/2))

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fricas [A]  time = 1.78, size = 540, normalized size = 2.90 \begin {gather*} \left [\frac {15 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d + 73 \, a b^{3} c^{2} d^{2} - 55 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (17 \, b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (59 \, b^{4} c^{2} d^{2} + 18 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{4} d^{2}}, \frac {15 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d + 73 \, a b^{3} c^{2} d^{2} - 55 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (17 \, b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (59 \, b^{4} c^{2} d^{2} + 18 \, a b^{3} c d^{3} - 5 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{4} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/768*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(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)*sqrt(d*x + c) + 8*(b^2*c*d
+ a*b*d^2)*x) + 4*(48*b^4*d^4*x^3 + 15*b^4*c^3*d + 73*a*b^3*c^2*d^2 - 55*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(17*
b^4*c*d^3 + a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 18*a*b^3*c*d^3 - 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(b^4*d^2), 1/384*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(-b*d)*arc
tan(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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d + 73*a*b^3*c^2*d^2 - 55*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(17*b^4*c*
d^3 + a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 18*a*b^3*c*d^3 - 5*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^
4*d^2)]

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giac [B]  time = 2.51, size = 1083, normalized size = 5.82

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-1/192*(192*((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*c^2*abs(b)/b^2 - 16*(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))*c*d*abs(b)/b - 8*(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*d^2*abs(b)/b
^2 - (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))*d^2*abs(b)/b - 48*(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) + sq
rt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*c^2*abs(b)/b^2 - 96*(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)*s
qrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a*c*d*abs(b)/b^3)/b

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maple [B]  time = 0.01, size = 641, normalized size = 3.45 \begin {gather*} -\frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{4} d^{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}\, b^{3}}+\frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} c \,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 )}{32 \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^{2} c^{2} 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 )}{64 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b}+\frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,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 )}{32 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}-\frac {5 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,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 )}{128 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d}+\frac {5 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} d^{2}}{64 b^{3}}-\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} c d}{64 b^{2}}+\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,c^{2}}{64 b}-\frac {5 \sqrt {d x +c}\, \sqrt {b x +a}\, c^{3}}{64 d}-\frac {5 \left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a^{2} d}{96 b^{2}}+\frac {5 \left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a c}{48 b}-\frac {5 \left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, c^{2}}{96 d}+\frac {\left (d x +c \right )^{\frac {5}{2}} \sqrt {b x +a}\, a}{24 b}-\frac {\left (d x +c \right )^{\frac {5}{2}} \sqrt {b x +a}\, c}{24 d}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {7}{2}}}{4 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(d*x+c)^(5/2),x)

[Out]

1/4/d*(b*x+a)^(1/2)*(d*x+c)^(7/2)+1/24/b*(d*x+c)^(5/2)*(b*x+a)^(1/2)*a-1/24/d*(d*x+c)^(5/2)*(b*x+a)^(1/2)*c-5/
96*d/b^2*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a^2+5/48/b*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a*c-5/96/d*(d*x+c)^(3/2)*(b*x+a)^(
1/2)*c^2+5/64*d^2/b^3*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^3-15/64*d/b^2*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2*c+15/64/b*(d
*x+c)^(1/2)*(b*x+a)^(1/2)*a*c^2-5/64/d*(d*x+c)^(1/2)*(b*x+a)^(1/2)*c^3-5/128*d^3/b^3*((b*x+a)*(d*x+c))^(1/2)/(
d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)
*a^4+5/32*d^2/b^2*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)*a^3*c-15/64*d/b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2
)*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))/(b*d)^(1/2)*a^2*c^2+5/32*((b*x+a)*(d
*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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
))/(b*d)^(1/2)*a*c^3-5/128/d*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)*c^4*b

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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)^(1/2)*(d*x+c)^(5/2),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 = -0.01 \begin {gather*} \int \sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)*(c + d*x)^(5/2),x)

[Out]

int((a + b*x)^(1/2)*(c + d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(d*x+c)**(5/2),x)

[Out]

Timed out

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