Optimal. Leaf size=140 \[ \frac {15 d^2}{4 \sqrt {c+d x} (b c-a d)^3}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac {5 d}{4 (a+b x) \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)} \]
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Rubi [A] time = 0.05, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 63, 208} \begin {gather*} \frac {15 d^2}{4 \sqrt {c+d x} (b c-a d)^3}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}+\frac {5 d}{4 (a+b x) \sqrt {c+d x} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 208
Rubi steps
\begin {align*} \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx &=-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}-\frac {(5 d) \int \frac {1}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{4 (b c-a d)}\\ &=-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {\left (15 d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^2}\\ &=\frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {\left (15 b d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 (b c-a d)^3}\\ &=\frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {(15 b d) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 (b c-a d)^3}\\ &=\frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.01, size = 50, normalized size = 0.36 \begin {gather*} -\frac {2 d^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{\sqrt {c+d x} (a d-b c)^3} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.49, size = 163, normalized size = 1.16 \begin {gather*} \frac {d^2 \left (8 a^2 d^2+25 a b d (c+d x)-16 a b c d+8 b^2 c^2+15 b^2 (c+d x)^2-25 b^2 c (c+d x)\right )}{4 \sqrt {c+d x} (b c-a d)^3 (-a d-b (c+d x)+b c)^2}+\frac {15 \sqrt {b} d^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{4 (a d-b c)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.28, size = 782, normalized size = 5.59 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {{\left (b c - a d\right )} \sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}}{b d x + b c}\right ) - {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.03, size = 234, normalized size = 1.67 \begin {gather*} \frac {15 \, b d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {d x + c}} + \frac {7 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 9 \, \sqrt {d x + c} b^{2} c d^{2} + 9 \, \sqrt {d x + c} a b d^{3}}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 179, normalized size = 1.28 \begin {gather*} -\frac {9 \sqrt {d x +c}\, a b \,d^{3}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}+\frac {9 \sqrt {d x +c}\, b^{2} c \,d^{2}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}-\frac {7 \left (d x +c \right )^{\frac {3}{2}} b^{2} d^{2}}{4 \left (a d -b c \right )^{3} \left (b d x +a d \right )^{2}}-\frac {15 b \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 \left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {2 d^{2}}{\left (a d -b c \right )^{3} \sqrt {d x +c}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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mupad [B] time = 0.44, size = 205, normalized size = 1.46 \begin {gather*} -\frac {\frac {2\,d^2}{a\,d-b\,c}+\frac {15\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{4\,{\left (a\,d-b\,c\right )}^3}+\frac {25\,b\,d^2\,\left (c+d\,x\right )}{4\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{5/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{3/2}+\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {15\,\sqrt {b}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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