3.15.23 \(\int \frac {1}{(a+b x)^5 \sqrt {c+d x}} \, dx\) [1423]

Optimal. Leaf size=180 \[ -\frac {\sqrt {c+d x}}{4 (b c-a d) (a+b x)^4}+\frac {7 d \sqrt {c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac {35 d^2 \sqrt {c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac {35 d^3 \sqrt {c+d x}}{64 (b c-a d)^4 (a+b x)}-\frac {35 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 \sqrt {b} (b c-a d)^{9/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {44, 65, 214} \begin {gather*} -\frac {35 d^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 \sqrt {b} (b c-a d)^{9/2}}+\frac {35 d^3 \sqrt {c+d x}}{64 (a+b x) (b c-a d)^4}-\frac {35 d^2 \sqrt {c+d x}}{96 (a+b x)^2 (b c-a d)^3}+\frac {7 d \sqrt {c+d x}}{24 (a+b x)^3 (b c-a d)^2}-\frac {\sqrt {c+d x}}{4 (a+b x)^4 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^5*Sqrt[c + d*x]),x]

[Out]

-1/4*Sqrt[c + d*x]/((b*c - a*d)*(a + b*x)^4) + (7*d*Sqrt[c + d*x])/(24*(b*c - a*d)^2*(a + b*x)^3) - (35*d^2*Sq
rt[c + d*x])/(96*(b*c - a*d)^3*(a + b*x)^2) + (35*d^3*Sqrt[c + d*x])/(64*(b*c - a*d)^4*(a + b*x)) - (35*d^4*Ar
cTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*Sqrt[b]*(b*c - a*d)^(9/2))

Rule 44

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

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]
time = 0.52, size = 166, normalized size = 0.92 \begin {gather*} \frac {1}{192} \left (\frac {\sqrt {c+d x} \left (279 a^3 d^3+a^2 b d^2 (-326 c+511 d x)+a b^2 d \left (200 c^2-252 c d x+385 d^2 x^2\right )+b^3 \left (-48 c^3+56 c^2 d x-70 c d^2 x^2+105 d^3 x^3\right )\right )}{(b c-a d)^4 (a+b x)^4}+\frac {105 d^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} (-b c+a d)^{9/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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[1/((a + b*x)^5*(c + d*x)^(1/2)),x]')

[Out]

Timed out

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Maple [A]
time = 0.16, size = 236, normalized size = 1.31

method	result	size
derivativedivides	\(2 d^{4} \left (\frac {\sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {\frac {7 \sqrt {d x +c}}{48 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {7 \left (\frac {5 \sqrt {d x +c}}{24 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {5 \left (\frac {3 \sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}\right )}{6 \left (a d -b c \right )}\right )}{8 \left (a d -b c \right )}}{a d -b c}\right )\)	\(236\)
default	\(2 d^{4} \left (\frac {\sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{4}}+\frac {\frac {7 \sqrt {d x +c}}{48 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {7 \left (\frac {5 \sqrt {d x +c}}{24 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {5 \left (\frac {3 \sqrt {d x +c}}{8 \left (a d -b c \right ) \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {3 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}\right )}{6 \left (a d -b c \right )}\right )}{8 \left (a d -b c \right )}}{a d -b c}\right )\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*(1/8*(d*x+c)^(1/2)/(a*d-b*c)/((d*x+c)*b+a*d-b*c)^4+7/8/(a*d-b*c)*(1/6*(d*x+c)^(1/2)/(a*d-b*c)/((d*x+c)*b
+a*d-b*c)^3+5/6/(a*d-b*c)*(1/4*(d*x+c)^(1/2)/(a*d-b*c)/((d*x+c)*b+a*d-b*c)^2+3/4/(a*d-b*c)*(1/2*(d*x+c)^(1/2)/
(a*d-b*c)/((d*x+c)*b+a*d-b*c)+1/2/(a*d-b*c)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(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(1/(b*x+a)^5/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (152) = 304\).
time = 0.32, size = 1325, normalized size = 7.36

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(b^2*c - a*b*d)*
log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(48*b^5*c^4 - 248*a*b^4*c^3*d +
526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d^2
- 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2*d
^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c*d
^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5*b
^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a^6
*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4 -
a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c*d
^4 - a^8*b^2*d^5)*x), 1/192*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)
*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (48*b^5*c^4 - 248*a*b^4*c^3*d
+ 526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d
^2 - 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2
*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c
*d^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5
*b^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a
^6*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4
- a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c
*d^4 - a^8*b^2*d^5)*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(1/(b*x+a)**5/(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. 331 vs. \(2 (152) = 304\).
time = 0.00, size = 409, normalized size = 2.27 \begin {gather*} 2 \left (\frac {105 \sqrt {c+d x} \left (c+d x\right )^{3} b^{3} d^{4}-385 \sqrt {c+d x} \left (c+d x\right )^{2} b^{3} d^{4} c+385 \sqrt {c+d x} \left (c+d x\right )^{2} b^{2} d^{5} a+511 \sqrt {c+d x} \left (c+d x\right ) b^{3} d^{4} c^{2}-1022 \sqrt {c+d x} \left (c+d x\right ) b^{2} d^{5} c a+511 \sqrt {c+d x} \left (c+d x\right ) b d^{6} a^{2}-279 \sqrt {c+d x} b^{3} d^{4} c^{3}+837 \sqrt {c+d x} b^{2} d^{5} c^{2} a-837 \sqrt {c+d x} b d^{6} c a^{2}+279 \sqrt {c+d x} d^{7} a^{3}}{\left (384 b^{4} c^{4}-1536 b^{3} d c^{3} a+2304 b^{2} d^{2} c^{2} a^{2}-1536 b d^{3} c a^{3}+384 d^{4} a^{4}\right ) \left (\left (c+d x\right ) b-b c+d a\right )^{4}}+\frac {35 d^{4} \arctan \left (\frac {b \sqrt {c+d x}}{\sqrt {-b^{2} c+a b d}}\right )}{2 \left (64 b^{4} c^{4}-256 b^{3} d c^{3} a+384 b^{2} d^{2} c^{2} a^{2}-256 b d^{3} c a^{3}+64 d^{4} a^{4}\right ) \sqrt {-b^{2} c+a b d}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((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^2*c + a*b*d)) + 1/192*(105*(d*x + c)^(7/2)*b^3*d^4 - 385*(d*x + c)^(5/2)*b^3*c*d^4 +
511*(d*x + c)^(3/2)*b^3*c^2*d^4 - 279*sqrt(d*x + c)*b^3*c^3*d^4 + 385*(d*x + c)^(5/2)*a*b^2*d^5 - 1022*(d*x +
c)^(3/2)*a*b^2*c*d^5 + 837*sqrt(d*x + c)*a*b^2*c^2*d^5 + 511*(d*x + c)^(3/2)*a^2*b*d^6 - 837*sqrt(d*x + c)*a^
2*b*c*d^6 + 279*sqrt(d*x + c)*a^3*d^7)/((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
)*((d*x + c)*b - b*c + a*d)^4)

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Mupad [B]
time = 0.46, size = 307, normalized size = 1.71 \begin {gather*} \frac {\frac {93\,d^4\,\sqrt {c+d\,x}}{64\,\left (a\,d-b\,c\right )}+\frac {385\,b^2\,d^4\,{\left (c+d\,x\right )}^{5/2}}{192\,{\left (a\,d-b\,c\right )}^3}+\frac {35\,b^3\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^4}+\frac {511\,b\,d^4\,{\left (c+d\,x\right )}^{3/2}}{192\,{\left (a\,d-b\,c\right )}^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3}+\frac {35\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,\sqrt {b}\,{\left (a\,d-b\,c\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((93*d^4*(c + d*x)^(1/2))/(64*(a*d - b*c)) + (385*b^2*d^4*(c + d*x)^(5/2))/(192*(a*d - b*c)^3) + (35*b^3*d^4*(
c + d*x)^(7/2))/(64*(a*d - b*c)^4) + (511*b*d^4*(c + d*x)^(3/2))/(192*(a*d - b*c)^2))/(b^4*(c + d*x)^4 - (4*b^
4*c - 4*a*b^3*d)*(c + d*x)^3 - (c + d*x)*(4*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^2 - 12*a*b^3*c^2*d) + a^4*d
^4 + b^4*c^4 + (c + d*x)^2*(6*b^4*c^2 + 6*a^2*b^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*
a^3*b*c*d^3) + (35*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(1/2)*(a*d - b*c)^(9/2))

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