Optimal. Leaf size=167 \[ -\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}+\frac {35 b d^2}{4 \sqrt {c+d x} (b c-a d)^4}+\frac {35 d^2}{12 (c+d x)^{3/2} (b c-a d)^3}+\frac {7 d}{4 (a+b x) (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (c+d x)^{3/2} (b c-a d)} \]
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Rubi [A] time = 0.06, antiderivative size = 167, 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 = {51, 63, 208} \[ -\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}+\frac {35 b d^2}{4 \sqrt {c+d x} (b c-a d)^4}+\frac {35 d^2}{12 (c+d x)^{3/2} (b c-a d)^3}+\frac {7 d}{4 (a+b x) (c+d x)^{3/2} (b c-a d)^2}-\frac {1}{2 (a+b x)^2 (c+d x)^{3/2} (b c-a d)} \]
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)^{5/2}} \, dx &=-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}-\frac {(7 d) \int \frac {1}{(a+b x)^2 (c+d x)^{5/2}} \, dx}{4 (b c-a d)}\\ &=-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{5/2}} \, dx}{8 (b c-a d)^2}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {\left (35 b d^2\right ) \int \frac {1}{(a+b x) (c+d x)^{3/2}} \, dx}{8 (b c-a d)^3}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{8 (b c-a d)^4}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}+\frac {\left (35 b^2 d\right ) \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)^4}\\ &=\frac {35 d^2}{12 (b c-a d)^3 (c+d x)^{3/2}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{3/2}}+\frac {7 d}{4 (b c-a d)^2 (a+b x) (c+d x)^{3/2}}+\frac {35 b d^2}{4 (b c-a d)^4 \sqrt {c+d x}}-\frac {35 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{9/2}}\\ \end {align*}
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Mathematica [C] time = 0.02, size = 52, normalized size = 0.31 \[ -\frac {2 d^2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {b (c+d x)}{a d-b c}\right )}{3 (c+d x)^{3/2} (a d-b c)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.52, size = 1226, normalized size = 7.34 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.20, size = 298, normalized size = 1.78 \[ \frac {35 \, b^{2} d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (9 \, {\left (d x + c\right )} b d^{2} + b c d^{2} - a d^{3}\right )}}{3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} d^{2} - 13 \, \sqrt {d x + c} b^{3} c d^{2} + 13 \, \sqrt {d x + c} a b^{2} d^{3}}{4 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 206, normalized size = 1.23 \[ \frac {13 \sqrt {d x +c}\, a \,b^{2} d^{3}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}-\frac {13 \sqrt {d x +c}\, b^{3} c \,d^{2}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}+\frac {11 \left (d x +c \right )^{\frac {3}{2}} b^{3} d^{2}}{4 \left (a d -b c \right )^{4} \left (b d x +a d \right )^{2}}+\frac {35 b^{2} d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{4 \left (a d -b c \right )^{4} \sqrt {\left (a d -b c \right ) b}}+\frac {6 b \,d^{2}}{\left (a d -b c \right )^{4} \sqrt {d x +c}}-\frac {2 d^{2}}{3 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.28, size = 243, normalized size = 1.46 \[ \frac {\frac {175\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{12\,{\left (a\,d-b\,c\right )}^3}-\frac {2\,d^2}{3\,\left (a\,d-b\,c\right )}+\frac {35\,b^3\,d^2\,{\left (c+d\,x\right )}^3}{4\,{\left (a\,d-b\,c\right )}^4}+\frac {14\,b\,d^2\,\left (c+d\,x\right )}{3\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{7/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{5/2}+{\left (c+d\,x\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {35\,b^{3/2}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4\right )}{{\left (a\,d-b\,c\right )}^{9/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{9/2}} \]
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 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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