Optimal. Leaf size=111 \[ -\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Shi}(a+b x)}{2 x^2}-\frac {b \sinh (a+b x)}{2 a x} \]
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Rubi [A] time = 0.35, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6532, 6742, 3297, 3303, 3298, 3301} \[ -\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Shi}(a+b x)}{2 x^2}-\frac {b \sinh (a+b x)}{2 a x} \]
Antiderivative was successfully verified.
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Rule 3297
Rule 3298
Rule 3301
Rule 3303
Rule 6532
Rule 6742
Rubi steps
\begin {align*} \int \frac {\text {Shi}(a+b x)}{x^3} \, dx &=-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {1}{2} b \int \frac {\sinh (a+b x)}{x^2 (a+b x)} \, dx\\ &=-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {1}{2} b \int \left (\frac {\sinh (a+b x)}{a x^2}-\frac {b \sinh (a+b x)}{a^2 x}+\frac {b^2 \sinh (a+b x)}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {b \int \frac {\sinh (a+b x)}{x^2} \, dx}{2 a}-\frac {b^2 \int \frac {\sinh (a+b x)}{x} \, dx}{2 a^2}+\frac {b^3 \int \frac {\sinh (a+b x)}{a+b x} \, dx}{2 a^2}\\ &=-\frac {b \sinh (a+b x)}{2 a x}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {b^2 \int \frac {\cosh (a+b x)}{x} \, dx}{2 a}-\frac {\left (b^2 \cosh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx}{2 a^2}-\frac {\left (b^2 \sinh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx}{2 a^2}\\ &=-\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2}+\frac {\left (b^2 \cosh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx}{2 a}+\frac {\left (b^2 \sinh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx}{2 a}\\ &=\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}-\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 86, normalized size = 0.77 \[ \frac {a^2 (-\text {Shi}(a+b x))+b^2 x^2 (a \cosh (a)-\sinh (a)) \text {Chi}(b x)+b^2 x^2 \text {Shi}(a+b x)+b^2 x^2 (a \sinh (a)-\cosh (a)) \text {Shi}(b x)-a b x \sinh (a+b x)}{2 a^2 x^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {Shi}\left (b x + a\right )}{x^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Shi}\left (b x + a\right )}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\Shi \left (b x +a \right )}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Shi}\left (b x + a\right )}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinhint}\left (a+b\,x\right )}{x^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Shi}{\left (a + b x \right )}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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