Optimal. Leaf size=147 \[ \frac {i \cosh (2 c+2 d x) \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^m \left (\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-i \sinh (2 c+2 d x)),\frac {b (1-i \sinh (2 c+2 d x))}{2 i a+b}\right )}{\sqrt {2} d \sqrt {1+i \sinh (2 c+2 d x)}} \]
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Rubi [A] time = 0.13, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2666, 2665, 139, 138} \[ \frac {i \cosh (2 c+2 d x) \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^m \left (\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-i \sinh (2 c+2 d x)),\frac {b (1-i \sinh (2 c+2 d x))}{2 i a+b}\right )}{\sqrt {2} d \sqrt {1+i \sinh (2 c+2 d x)}} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 2665
Rule 2666
Rubi steps
\begin {align*} \int (a+b \cosh (c+d x) \sinh (c+d x))^m \, dx &=\int \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^m \, dx\\ &=-\frac {(i \cosh (2 c+2 d x)) \operatorname {Subst}\left (\int \frac {\left (a-\frac {i b x}{2}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,i \sinh (2 c+2 d x)\right )}{2 d \sqrt {1-i \sinh (2 c+2 d x)} \sqrt {1+i \sinh (2 c+2 d x)}}\\ &=-\frac {\left (i \cosh (2 c+2 d x) \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^m \left (-\frac {a+\frac {1}{2} b \sinh (2 c+2 d x)}{-a+\frac {i b}{2}}\right )^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{-a+\frac {i b}{2}}+\frac {i b x}{2 \left (-a+\frac {i b}{2}\right )}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,i \sinh (2 c+2 d x)\right )}{2 d \sqrt {1-i \sinh (2 c+2 d x)} \sqrt {1+i \sinh (2 c+2 d x)}}\\ &=\frac {i F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-i \sinh (2 c+2 d x)),\frac {b (1-i \sinh (2 c+2 d x))}{2 i a+b}\right ) \cosh (2 c+2 d x) \left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^m \left (\frac {2 a+b \sinh (2 c+2 d x)}{2 a-i b}\right )^{-m}}{\sqrt {2} d \sqrt {1+i \sinh (2 c+2 d x)}}\\ \end {align*}
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Mathematica [A] time = 0.69, size = 162, normalized size = 1.10 \[ \frac {\text {sech}(2 (c+d x)) \sqrt {\frac {b (1-i \sinh (2 (c+d x)))}{b+2 i a}} \sqrt {\frac {b (1+i \sinh (2 (c+d x)))}{b-2 i a}} \left (a+\frac {1}{2} b \sinh (2 (c+d x))\right )^{m+1} F_1\left (m+1;\frac {1}{2},\frac {1}{2};m+2;\frac {2 a+b \sinh (2 (c+d x))}{2 a+i b},\frac {2 a+b \sinh (2 (c+d x))}{2 a-i b}\right )}{b d (m+1)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{m}, x\right ) \]
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 \[ \int {\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.02, size = 0, normalized size = 0.00 \[ \int \left (a +b \cosh \left (d x +c \right ) \sinh \left (d x +c \right )\right )^{m}\, dx \]
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 \[ \int {\left (b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
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 \[ \int {\left (a+b\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )\right )}^m \,d x \]
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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