3.15.92 \(\int \frac {(a+b x)^{7/2}}{\sqrt {c+d x}} \, dx\) [1492]

Optimal. Leaf size=183 \[ -\frac {35 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 d^4}+\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {35 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 \sqrt {b} d^{9/2}} \]

[Out]

35/64*(-a*d+b*c)^4*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/d^(9/2)/b^(1/2)+35/96*(-a*d+b*c)^2*(b*
x+a)^(3/2)*(d*x+c)^(1/2)/d^3-7/24*(-a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d^2+1/4*(b*x+a)^(7/2)*(d*x+c)^(1/2)/d
-35/64*(-a*d+b*c)^3*(b*x+a)^(1/2)*(d*x+c)^(1/2)/d^4

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Rubi [A]
time = 0.07, antiderivative size = 183, 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 {35 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 \sqrt {b} d^{9/2}}-\frac {35 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 d^4}+\frac {35 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2}{96 d^3}-\frac {7 (a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/2)/Sqrt[c + d*x],x]

[Out]

(-35*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*d^4) + (35*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(9
6*d^3) - (7*(b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*d^2) + ((a + b*x)^(7/2)*Sqrt[c + d*x])/(4*d) + (35*
(b*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*Sqrt[b]*d^(9/2))

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 \frac {(a+b x)^{7/2}}{\sqrt {c+d x}} \, dx &=\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {(7 (b c-a d)) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx}{8 d}\\ &=-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {\left (35 (b c-a d)^2\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{48 d^2}\\ &=\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}-\frac {\left (35 (b c-a d)^3\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{64 d^3}\\ &=-\frac {35 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 d^4}+\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {\left (35 (b c-a d)^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 d^4}\\ &=-\frac {35 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 d^4}+\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {\left (35 (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 d^4}\\ &=-\frac {35 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 d^4}+\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {\left (35 (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 d^4}\\ &=-\frac {35 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 d^4}+\frac {35 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{96 d^3}-\frac {7 (b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{24 d^2}+\frac {(a+b x)^{7/2} \sqrt {c+d x}}{4 d}+\frac {35 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 \sqrt {b} d^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 164, normalized size = 0.90 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (279 a^3 d^3+a^2 b d^2 (-511 c+326 d x)+a b^2 d \left (385 c^2-252 c d x+200 d^2 x^2\right )+b^3 \left (-105 c^3+70 c^2 d x-56 c d^2 x^2+48 d^3 x^3\right )\right )}{192 d^4}+\frac {35 (b c-a d)^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 \sqrt {b} d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(279*a^3*d^3 + a^2*b*d^2*(-511*c + 326*d*x) + a*b^2*d*(385*c^2 - 252*c*d*x + 200*
d^2*x^2) + b^3*(-105*c^3 + 70*c^2*d*x - 56*c*d^2*x^2 + 48*d^3*x^3)))/(192*d^4) + (35*(b*c - a*d)^4*ArcTanh[(Sq
rt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(64*Sqrt[b]*d^(9/2))

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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)^(7/2)/(c + d*x)^(1/2),x]')

[Out]

Timed out

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Maple [A]
time = 0.16, size = 206, normalized size = 1.13

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {7}{2}} \sqrt {d x +c}}{4 d}-\frac {7 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}{3 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}{2 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{d}-\frac {\left (-a d +b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 d \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}\right )}{4 d}\right )}{6 d}\right )}{8 d}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 542, normalized size = 2.96 \begin {gather*} \left [\frac {105 \, {\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} - 105 \, b^{4} c^{3} d + 385 \, a b^{3} c^{2} d^{2} - 511 \, a^{2} b^{2} c d^{3} + 279 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 25 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 126 \, a b^{3} c d^{3} + 163 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, b d^{5}}, -\frac {105 \, {\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} - 105 \, b^{4} c^{3} d + 385 \, a b^{3} c^{2} d^{2} - 511 \, a^{2} b^{2} c d^{3} + 279 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 25 \, a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 126 \, a b^{3} c d^{3} + 163 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, b 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]

[1/768*(105*(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 - 105*b^4*c^3*d + 385*a*b^3*c^2*d^2 - 511*a^2*b^2*c*d^3 + 279*a^3*b*d^4 - 8
*(7*b^4*c*d^3 - 25*a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 126*a*b^3*c*d^3 + 163*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sq
rt(d*x + c))/(b*d^5), -1/384*(105*(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)*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*(48*b^4*d^4*x^3 - 105*b^4*c^3*d + 385*a*b^3*c^2*d^2 - 511*a^2*b^2*c*d^3 + 279*a^3*b*d^4
- 8*(7*b^4*c*d^3 - 25*a*b^3*d^4)*x^2 + 2*(35*b^4*c^2*d^2 - 126*a*b^3*c*d^3 + 163*a^2*b^2*d^4)*x)*sqrt(b*x + a)
*sqrt(d*x + c))/(b*d^5)]

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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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Giac [A]
time = 0.02, size = 353, normalized size = 1.93 \begin {gather*} \frac {b^{2} \left (2 \left (\left (\left (\frac {\frac {1}{5760}\cdot 720 d^{6} \sqrt {a+b x} \sqrt {a+b x}}{b d^{7}}-\frac {\frac {1}{5760} \left (840 b d^{5} c-840 d^{6} a\right )}{b d^{7}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{5760} \left (-1050 b^{2} d^{4} c^{2}+2100 b d^{5} a c-1050 d^{6} a^{2}\right )}{b d^{7}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{5760} \left (1575 b^{3} d^{3} c^{3}-4725 b^{2} d^{4} a c^{2}+4725 b d^{5} a^{2} c-1575 d^{6} a^{3}\right )}{b d^{7}}\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}+140 a^{3} b c d^{3}-210 a^{2} b^{2} c^{2} d^{2}+140 a b^{3} c^{3} d-35 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 |}{128 d^{4} \sqrt {b d}}\right )}{\left |b\right | b} \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]

1/192*(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*d) - 7*(b*c*d^5 - a*d^6)/
(b*d^7)) + 35*(b^2*c^2*d^4 - 2*a*b*c*d^5 + a^2*d^6)/(b*d^7)) - 105*(b^3*c^3*d^3 - 3*a*b^2*c^2*d^4 + 3*a^2*b*c*
d^5 - a^3*d^6)/(b*d^7))*sqrt(b*x + a) - 105*(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)*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^4))*b/abs(b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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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