Optimal. Leaf size=295 \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}+b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac {x}{2 b} \]
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Rubi [A] time = 0.56, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3220, 2635, 8, 2660, 618, 206, 204} \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}+b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac {x}{2 b} \]
Antiderivative was successfully verified.
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Rule 8
Rule 204
Rule 206
Rule 618
Rule 2635
Rule 2660
Rule 3220
Rubi steps
\begin {align*} \int \frac {\sinh ^5(c+d x)}{a+b \sinh ^3(c+d x)} \, dx &=-\left (i \int \left (\frac {i \sinh ^2(c+d x)}{b}-\frac {i a \sinh ^2(c+d x)}{b \left (a+b \sinh ^3(c+d x)\right )}\right ) \, dx\right )\\ &=\frac {\int \sinh ^2(c+d x) \, dx}{b}-\frac {a \int \frac {\sinh ^2(c+d x)}{a+b \sinh ^3(c+d x)} \, dx}{b}\\ &=\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac {\int 1 \, dx}{2 b}+\frac {a \int \left (\frac {i}{3 b^{2/3} \left (-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac {i}{3 b^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac {i}{3 b^{2/3} \left ((-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}\right ) \, dx}{b}\\ &=-\frac {x}{2 b}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d}+\frac {(i a) \int \frac {1}{-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{5/3}}+\frac {(i a) \int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{5/3}}+\frac {(i a) \int \frac {1}{(-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 b^{5/3}}\\ &=-\frac {x}{2 b}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-i \sqrt [3]{a}-2 \sqrt [3]{b} x-i \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-2 \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{(-1)^{5/6} \sqrt [3]{a}-2 \sqrt [3]{b} x+(-1)^{5/6} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}\\ &=-\frac {x}{2 b}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{-4 \left (\sqrt [3]{-1} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}-\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}-2 i \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}-\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{4 \left ((-1)^{2/3} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 (-1)^{5/6} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 b^{5/3} d}\\ &=-\frac {x}{2 b}-\frac {2 a \tan ^{-1}\left (\frac {\sqrt [3]{b}+\sqrt [3]{-1} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}} b^{5/3} d}-\frac {2 a \tan ^{-1}\left (\frac {\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}} b^{5/3} d}+\frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {a^{2/3}+b^{2/3}} b^{5/3} d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d}\\ \end {align*}
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Mathematica [C] time = 0.33, size = 299, normalized size = 1.01 \[ \frac {-2 a \text {RootSum}\left [\text {$\#$1}^6 b-3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b-b\& ,\frac {2 \text {$\#$1}^4 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\text {$\#$1}^4 c+\text {$\#$1}^4 d x-4 \text {$\#$1}^2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )-2 \text {$\#$1}^2 c-2 \text {$\#$1}^2 d x+2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+c+d x}{\text {$\#$1}^5 b-2 \text {$\#$1}^3 b+4 \text {$\#$1}^2 a+\text {$\#$1} b}\& \right ]-6 (c+d x)+3 \sinh (2 (c+d x))}{12 b d} \]
Antiderivative was successfully verified.
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fricas [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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (d x + c\right )^{5}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.11, size = 207, normalized size = 0.70 \[ \frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d b}-\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d b}+\frac {4 a \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d b} \]
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 \[ -\frac {{\left (4 \, d x e^{\left (2 \, d x + 2 \, c\right )} - e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b d} - \frac {1}{32} \, \int \frac {64 \, {\left (a e^{\left (5 \, d x + 5 \, c\right )} - 2 \, a e^{\left (3 \, d x + 3 \, c\right )} + a e^{\left (d x + c\right )}\right )}}{b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.48, size = 1114, normalized size = 3.78 \[ \left (\sum _{k=1}^6\ln \left (-\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\frac {\left (8\,a^6\,d^4+4\,a^4\,b^2\,d^4-a^5\,b\,d^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,663552}{b^7}+\frac {\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (-a^4\,b\,d^5+a^5\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,4+a^3\,b^2\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,1990656}{b^5}\right )+\frac {\left (4\,a^6\,b\,d^3+a^7\,d^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,8-a^5\,b^2\,d^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,442368}{b^9}\right )-\frac {a^6\,d^2\,\left (2\,b-a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,294912}{b^{10}}\right )-\frac {a^7\,d\,\left (8\,a-b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,24576}{b^{12}}\right )-\frac {a^8\,\left (b-a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,4\right )\,32768}{b^{14}}\right )\,\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\right )-\frac {x}{2\,b}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d} \]
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 \[ \int \frac {\sinh ^{5}{\left (c + d x \right )}}{a + b \sinh ^{3}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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