3.15.61 \(\int (a+b x)^{5/2} \sqrt {c+d x} \, dx\) [1461]

Optimal. Leaf size=192 \[ \frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}-\frac {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 192, 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 = {52, 65, 223, 212} \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^{3/2} d^{7/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 b d^3}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2}{96 b d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)*Sqrt[c + d*x],x]

[Out]

(5*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^3) - (5*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(96
*b*d^2) + ((b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*b*d) + ((a + b*x)^(7/2)*Sqrt[c + d*x])/(4*b) - (5*(b
*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(7/2))

Rule 52

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 65

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 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]

Rubi steps

\begin {align*} \int (a+b x)^{5/2} \sqrt {c+d x} \, dx &=\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}+\frac {(b c-a d) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{8 b}\\ &=\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}-\frac {\left (5 (b c-a d)^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{48 b d}\\ &=-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}+\frac {\left (5 (b c-a d)^3\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 b d^2}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{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 d^3}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}-\frac {\left (5 (b c-a d)^4\right ) \text {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^2 d^3}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 b}-\frac {\left (5 (b c-a d)^4\right ) \text {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^2 d^3}\\ &=\frac {5 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b d^3}-\frac {5 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^2}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{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^{3/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 144, normalized size = 0.75 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 d^3 (a+b x)^3+73 b d^2 (a+b x)^2 (c+d x)-55 b^2 d (a+b x) (c+d x)^2+15 b^3 (c+d x)^3\right )}{192 b d^3}-\frac {5 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{3/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(a + b*x)^(5/2)*(c + d*x)^(1/2),x]')

[Out]

Timed out

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Maple [A]
time = 0.18, size = 206, normalized size = 1.07

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{2}}}{4 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}}}{2 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 d}\right )}{2 d}\right )}{8 d}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2),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 detail

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Fricas [A]
time = 0.32, size = 540, normalized size = 2.81 \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 - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b^{2} d^{4}}, \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 - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \, {\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b^{2} 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)^(1/2),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 - 55*a*b^3*c^2*d^2 + 73*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4
*c*d^3 + 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 18*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(b^2*d^4), 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 - 55*a*b^3*c^2*d^2 + 73*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4*c*d^3
+ 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 18*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^
2*d^4)]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 726 vs. \(2 (154) = 308\).
time = 0.07, size = 972, normalized size = 5.06 \begin {gather*} \frac {\frac {2 b^{3} \left |b\right | \left (2 \left (\left (\left (\frac {\frac {1}{184320}\cdot 11520 b^{11} d^{6} \sqrt {a+b x} \sqrt {a+b x}}{b^{14} d^{6}}-\frac {\frac {1}{184320} \left (-1920 b^{12} d^{5} c+48000 b^{11} d^{6} a\right )}{b^{14} d^{6}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{184320} \left (2400 b^{13} d^{4} c^{2}+6720 b^{12} d^{5} a c-78240 b^{11} d^{6} a^{2}\right )}{b^{14} d^{6}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{184320} \left (-3600 b^{14} d^{3} c^{3}-6480 b^{13} d^{4} a c^{2}-10800 b^{12} d^{5} a^{2} c+66960 b^{11} d^{6} a^{3}\right )}{b^{14} d^{6}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (-35 a^{4} d^{4}+20 a^{3} b c d^{3}+6 a^{2} b^{2} c^{2} d^{2}+4 a b^{3} c^{3} d+5 b^{4} c^{4}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{256 b^{2} d^{3} \sqrt {b d}}\right )}{b^{2}}+\frac {6 a b^{2} \left |b\right | \left (2 \left (\left (\frac {\frac {1}{2304}\cdot 192 b^{5} d^{4} \sqrt {a+b x} \sqrt {a+b x}}{b^{7} d^{4}}-\frac {\frac {1}{2304} \left (-48 b^{6} d^{3} c+624 b^{5} d^{4} a\right )}{b^{7} d^{4}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{2304} \left (72 b^{7} d^{2} c^{2}+144 b^{6} d^{3} a c-792 b^{5} d^{4} a^{2}\right )}{b^{7} d^{4}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (5 a^{3} d^{3}-3 a^{2} b c d^{2}-a b^{2} c^{2} d-b^{3} c^{3}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{32 b d^{2} \sqrt {b d}}\right )}{b^{2}}+\frac {6 a^{2} b \left |b\right | \left (2 \left (\frac {\frac {1}{64}\cdot 8 d^{2} \sqrt {a+b x} \sqrt {a+b x}}{d^{2}}-\frac {\frac {1}{64} \left (-4 b d c+20 d^{2} a\right )}{d^{2}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (-3 a^{2} b d^{2}+2 a b^{2} c d+b^{3} c^{2}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{16 d \sqrt {b d}}\right )}{b^{2} b}+\frac {2 a^{3} \left |b\right | \left (\frac {1}{2} \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (a b d-b^{2} c\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{4 \sqrt {b d}}\right )}{b^{2}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(24*(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*abs(b) - 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^3*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))*b*abs(b) + 144*(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^2*abs(b)/b^2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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