3.14.54 \(\int (a+b x)^{7/2} \sqrt {c+d x} \, dx\)

Optimal. Leaf size=230 \[ \frac {7 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{3/2} d^{9/2}}-\frac {7 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}{128 b d^4}+\frac {7 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}{192 b d^3}-\frac {7 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}{240 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b} \]

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Rubi [A]  time = 0.16, antiderivative size = 230, 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 {7 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{3/2} d^{9/2}}-\frac {7 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^4}{128 b d^4}+\frac {7 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^3}{192 b d^3}-\frac {7 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)^2}{240 b d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x} (b c-a d)}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/2)*Sqrt[c + d*x],x]

[Out]

(-7*(b*c - a*d)^4*Sqrt[a + b*x]*Sqrt[c + d*x])/(128*b*d^4) + (7*(b*c - a*d)^3*(a + b*x)^(3/2)*Sqrt[c + d*x])/(
192*b*d^3) - (7*(b*c - a*d)^2*(a + b*x)^(5/2)*Sqrt[c + d*x])/(240*b*d^2) + ((b*c - a*d)*(a + b*x)^(7/2)*Sqrt[c
+ d*x])/(40*b*d) + ((a + b*x)^(9/2)*Sqrt[c + d*x])/(5*b) + (7*(b*c - a*d)^5*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(
Sqrt[b]*Sqrt[c + d*x])])/(128*b^(3/2)*d^(9/2))

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)^{7/2} \sqrt {c+d x} \, dx &=\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {(b c-a d) \int \frac {(a+b x)^{7/2}}{\sqrt {c+d x}} \, dx}{10 b}\\ &=\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}-\frac {\left (7 (b c-a d)^2\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{80 b d}\\ &=-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {\left (7 (b c-a d)^3\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{96 b d^2}\\ &=\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}-\frac {\left (7 (b c-a d)^4\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{128 b d^3}\\ &=-\frac {7 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b d^4}+\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {\left (7 (b c-a d)^5\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{256 b d^4}\\ &=-\frac {7 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b d^4}+\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {\left (7 (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^2 d^4}\\ &=-\frac {7 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b d^4}+\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {\left (7 (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^2 d^4}\\ &=-\frac {7 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b d^4}+\frac {7 (b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{192 b d^3}-\frac {7 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{240 b d^2}+\frac {(b c-a d) (a+b x)^{7/2} \sqrt {c+d x}}{40 b d}+\frac {(a+b x)^{9/2} \sqrt {c+d x}}{5 b}+\frac {7 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{3/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 1.47, size = 194, normalized size = 0.84 \begin {gather*} \frac {(a+b x)^{9/2} \sqrt {c+d x} \left (\frac {70 (b c-a d)^{9/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{d^{9/2} (a+b x)^{9/2} \sqrt {\frac {b (c+d x)}{b c-a d}}}-\frac {70 (b c-a d)^4}{d^4 (a+b x)^4}+\frac {140 (b c-a d)^3}{3 d^3 (a+b x)^3}-\frac {112 (b c-a d)^2}{3 d^2 (a+b x)^2}+\frac {32 b c-32 a d}{a d+b d x}+256\right )}{1280 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(7/2)*Sqrt[c + d*x],x]

[Out]

((a + b*x)^(9/2)*Sqrt[c + d*x]*(256 - (70*(b*c - a*d)^4)/(d^4*(a + b*x)^4) + (140*(b*c - a*d)^3)/(3*d^3*(a + b
*x)^3) - (112*(b*c - a*d)^2)/(3*d^2*(a + b*x)^2) + (32*b*c - 32*a*d)/(a*d + b*d*x) + (70*(b*c - a*d)^(9/2)*Arc
Sinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(d^(9/2)*(a + b*x)^(9/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)])))/(12
80*b)

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IntegrateAlgebraic [A]  time = 0.31, size = 198, normalized size = 0.86 \begin {gather*} \frac {7 (b c-a d)^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{3/2} d^{9/2}}-\frac {\sqrt {c+d x} (b c-a d)^5 \left (\frac {105 b^4 (c+d x)^4}{(a+b x)^4}-\frac {490 b^3 d (c+d x)^3}{(a+b x)^3}+\frac {896 b^2 d^2 (c+d x)^2}{(a+b x)^2}-\frac {790 b d^3 (c+d x)}{a+b x}-105 d^4\right )}{1920 b d^4 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(7/2)*Sqrt[c + d*x],x]

[Out]

-1/1920*((b*c - a*d)^5*Sqrt[c + d*x]*(-105*d^4 - (790*b*d^3*(c + d*x))/(a + b*x) + (896*b^2*d^2*(c + d*x)^2)/(
a + b*x)^2 - (490*b^3*d*(c + d*x)^3)/(a + b*x)^3 + (105*b^4*(c + d*x)^4)/(a + b*x)^4))/(b*d^4*Sqrt[a + b*x]*(-
d + (b*(c + d*x))/(a + b*x))^5) + (7*(b*c - a*d)^5*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(
128*b^(3/2)*d^(9/2))

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fricas [A]  time = 1.74, size = 702, normalized size = 3.05 \begin {gather*} \left [-\frac {105 \, {\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 (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 490 \, a b^{4} c^{3} d^{2} - 896 \, a^{2} b^{3} c^{2} d^{3} + 790 \, a^{3} b^{2} c d^{4} + 105 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 31 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 32 \, a b^{4} c d^{4} - 263 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 161 \, a b^{4} c^{2} d^{3} + 289 \, a^{2} b^{3} c d^{4} + 605 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{7680 \, b^{2} d^{5}}, -\frac {105 \, {\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 (384 \, b^{5} d^{5} x^{4} - 105 \, b^{5} c^{4} d + 490 \, a b^{4} c^{3} d^{2} - 896 \, a^{2} b^{3} c^{2} d^{3} + 790 \, a^{3} b^{2} c d^{4} + 105 \, a^{4} b d^{5} + 48 \, {\left (b^{5} c d^{4} + 31 \, a b^{4} d^{5}\right )} x^{3} - 8 \, {\left (7 \, b^{5} c^{2} d^{3} - 32 \, a b^{4} c d^{4} - 263 \, a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (35 \, b^{5} c^{3} d^{2} - 161 \, a b^{4} c^{2} d^{3} + 289 \, a^{2} b^{3} c d^{4} + 605 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3840 \, b^{2} d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/7680*(105*(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)*sq
rt(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)*sq
rt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 490*a*b^4*c^3*d^2 - 896*a^2*b^3*
c^2*d^3 + 790*a^3*b^2*c*d^4 + 105*a^4*b*d^5 + 48*(b^5*c*d^4 + 31*a*b^4*d^5)*x^3 - 8*(7*b^5*c^2*d^3 - 32*a*b^4*
c*d^4 - 263*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 161*a*b^4*c^2*d^3 + 289*a^2*b^3*c*d^4 + 605*a^3*b^2*d^5)*x)
*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^5), -1/3840*(105*(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)*sqr
t(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) - 2*(384*b^5*d^5*x^4 - 105*b^5*c^4*d + 490*a*b^4*c
^3*d^2 - 896*a^2*b^3*c^2*d^3 + 790*a^3*b^2*c*d^4 + 105*a^4*b*d^5 + 48*(b^5*c*d^4 + 31*a*b^4*d^5)*x^3 - 8*(7*b^
5*c^2*d^3 - 32*a*b^4*c*d^4 - 263*a^2*b^3*d^5)*x^2 + 2*(35*b^5*c^3*d^2 - 161*a*b^4*c^2*d^3 + 289*a^2*b^3*c*d^4
+ 605*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d^5)]

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giac [B]  time = 1.84, size = 1107, normalized size = 4.81

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)*(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/1920*(480*(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^2*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^4*abs(b)/b^2 + 40*(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*b*abs(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))*s
qrt(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^2*abs(b
) + 1920*(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*abs(b)/b^2)/b

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maple [B]  time = 0.01, size = 858, normalized size = 3.73 \begin {gather*} -\frac {7 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{5} 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 {35 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{4} c \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}}-\frac {35 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} b \,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}\, d}+\frac {35 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} b^{2} 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^{2}}-\frac {35 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,b^{3} 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^{3}}+\frac {7 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{4} 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^{4}}+\frac {7 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{4}}{128 b}-\frac {7 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} c}{32 d}+\frac {21 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,c^{2}}{64 d^{2}}-\frac {7 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c^{3}}{32 d^{3}}+\frac {7 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{4}}{128 d^{4}}+\frac {7 \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} a^{3}}{64 d}-\frac {21 \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} a^{2} b c}{64 d^{2}}+\frac {21 \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} a \,b^{2} c^{2}}{64 d^{3}}-\frac {7 \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} b^{3} c^{3}}{64 d^{4}}+\frac {7 \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} a^{2}}{48 d}-\frac {7 \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} a b c}{24 d^{2}}+\frac {7 \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} b^{2} c^{2}}{48 d^{3}}+\frac {7 \left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{2}} a}{40 d}-\frac {7 \left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {3}{2}} b c}{40 d^{2}}+\frac {\left (b x +a \right )^{\frac {7}{2}} \left (d x +c \right )^{\frac {3}{2}}}{5 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/2)*(d*x+c)^(1/2),x)

[Out]

1/5/d*(b*x+a)^(7/2)*(d*x+c)^(3/2)+7/40/d*(b*x+a)^(5/2)*(d*x+c)^(3/2)*a+7/48/d*(b*x+a)^(3/2)*(d*x+c)^(3/2)*a^2+
7/64/d*(b*x+a)^(1/2)*(d*x+c)^(3/2)*a^3+21/64/d^3*(b*x+a)^(1/2)*(d*x+c)^(3/2)*a*b^2*c^2+21/64/d^2*(d*x+c)^(1/2)
*(b*x+a)^(1/2)*a^2*c^2*b-7/32/d^3*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*c^3*b^2-21/64/d^2*(b*x+a)^(1/2)*(d*x+c)^(3/2)*
a^2*b*c-7/24/d^2*(b*x+a)^(3/2)*(d*x+c)^(3/2)*a*b*c-7/40/d^2*(b*x+a)^(5/2)*(d*x+c)^(3/2)*b*c+7/48/d^3*(b*x+a)^(
3/2)*(d*x+c)^(3/2)*b^2*c^2-7/64/d^4*(b*x+a)^(1/2)*(d*x+c)^(3/2)*b^3*c^3+7/128/b*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^
4-7/32/d*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^3*c+7/128/d^4*(d*x+c)^(1/2)*(b*x+a)^(1/2)*c^4*b^3+35/256*((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)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/
(b*d)^(1/2)*a^4*c+35/128/d^2*((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)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^2*c^3*b^2-35/256/d^3*((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)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a*c^4*
b^3-7/256*d/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)+(d*x^
2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^5-35/128/d*((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)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^3*c^2*b+7/256/d^4*((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)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2
))/(b*d)^(1/2)*c^5*b^4

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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)^(7/2)*(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 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 )}^{7/2}\,\sqrt {c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(7/2)*(c + d*x)^(1/2),x)

[Out]

int((a + b*x)^(7/2)*(c + d*x)^(1/2), 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)**(7/2)*(d*x+c)**(1/2),x)

[Out]

Timed out

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