Integrand size = 18, antiderivative size = 220 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b^{5/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^{5/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}} \]
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Time = 0.23 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3395, 32, 3394, 12, 3389, 2211, 2235, 2236} \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {8 \sqrt {2 \pi } b^{5/2} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 \sqrt {2 \pi } b^{5/2} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}+\frac {16 b^2}{15 d^3 \sqrt {c+d x}} \]
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Rule 12
Rule 32
Rule 2211
Rule 2235
Rule 2236
Rule 3389
Rule 3394
Rule 3395
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {\left (8 b^2\right ) \int \frac {1}{(c+d x)^{3/2}} \, dx}{15 d^2}+\frac {\left (16 b^2\right ) \int \frac {\cosh ^2(a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2} \\ & = \frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (64 i b^3\right ) \int -\frac {i \sinh (2 a+2 b x)}{2 \sqrt {c+d x}} \, dx}{15 d^3} \\ & = \frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (32 b^3\right ) \int \frac {\sinh (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{15 d^3} \\ & = \frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {\left (16 b^3\right ) \int \frac {e^{-i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{15 d^3}-\frac {\left (16 b^3\right ) \int \frac {e^{i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{15 d^3} \\ & = \frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {\left (32 b^3\right ) \text {Subst}\left (\int e^{i \left (2 i a-\frac {2 i b c}{d}\right )-\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{15 d^4}+\frac {\left (32 b^3\right ) \text {Subst}\left (\int e^{-i \left (2 i a-\frac {2 i b c}{d}\right )+\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{15 d^4} \\ & = \frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {2 \cosh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \cosh ^2(a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {8 b^{5/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^{5/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.93 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {-6 d^2+e^{2 a} \left (-3 d^2 e^{2 b x}-4 b e^{-\frac {2 b c}{d}} (c+d x) \left (e^{\frac {2 b (c+d x)}{d}} (d+4 b (c+d x))+4 \sqrt {2} d \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )\right )\right )+e^{-2 (a+b x)} \left (-3 d^2+4 b (c+d x) \left (d-4 b (c+d x)+4 \sqrt {2} d e^{\frac {2 b (c+d x)}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )\right )\right )}{30 d^3 (c+d x)^{5/2}} \]
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\[\int \frac {\cosh \left (b x +a \right )^{2}}{\left (d x +c \right )^{\frac {7}{2}}}d x\]
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Leaf count of result is larger than twice the leaf count of optimal. 1350 vs. \(2 (172) = 344\).
Time = 0.31 (sec) , antiderivative size = 1350, normalized size of antiderivative = 6.14 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\cosh ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.53 \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {\frac {5 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {5}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} + \frac {5 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {5}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} + \frac {1}{{\left (d x + c\right )}^{\frac {5}{2}}}}{5 \, d} \]
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\[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int { \frac {\cosh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\cosh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \]
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