Optimal. Leaf size=258 \[ \frac{e^4 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^2 d}-\frac{9 e^4 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac{5 e^4 \sin \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac{e^4 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^2 d}+\frac{9 e^4 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac{5 e^4 \cos \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
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Rubi [A] time = 0.367686, antiderivative size = 254, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac{e^4 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}-\frac{9 e^4 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}+\frac{5 e^4 \sin \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{e^4 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{9 e^4 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{5 e^4 \cos \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4631
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^4}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^4 x^4}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^4 \operatorname{Subst}\left (\int \left (-\frac{\sin (x)}{8 (a+b x)}+\frac{9 \sin (3 x)}{16 (a+b x)}-\frac{5 \sin (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^4 \operatorname{Subst}\left (\int \frac{\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{\left (5 e^4\right ) \operatorname{Subst}\left (\int \frac{\sin (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}+\frac{\left (9 e^4\right ) \operatorname{Subst}\left (\int \frac{\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{\left (e^4 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}+\frac{\left (9 e^4 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}-\frac{\left (5 e^4 \cos \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}+\frac{\left (e^4 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}-\frac{\left (9 e^4 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}+\frac{\left (5 e^4 \sin \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac{e^4 (c+d x)^4 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^4 \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right ) \sin \left (\frac{a}{b}\right )}{8 b^2 d}-\frac{9 e^4 \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{3 a}{b}\right )}{16 b^2 d}+\frac{5 e^4 \text{Ci}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{5 a}{b}\right )}{16 b^2 d}-\frac{e^4 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{9 e^4 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac{5 e^4 \cos \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}\\ \end{align*}
Mathematica [A] time = 1.16289, size = 283, normalized size = 1.1 \[ \frac{e^4 \left (16 \left (-3 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+3 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )+5 \left (10 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )-5 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-10 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+5 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac{5 a}{b}\right ) \text{Si}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )-\frac{16 b \sqrt{1-(c+d x)^2} (c+d x)^4}{a+b \sin ^{-1}(c+d x)}\right )}{16 b^2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 396, normalized size = 1.5 \begin{align*} -{\frac{{e}^{4}}{16\,d \left ( a+b\arcsin \left ( dx+c \right ) \right ){b}^{2}} \left ( 5\,\arcsin \left ( dx+c \right ) \cos \left ( 5\,{\frac{a}{b}} \right ){\it Si} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) b-5\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) \sin \left ( 5\,{\frac{a}{b}} \right ) b+2\,\arcsin \left ( dx+c \right ){\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b-2\,\arcsin \left ( dx+c \right ){\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b-9\,\arcsin \left ( dx+c \right ){\it Si} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) b+9\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) b+5\,\cos \left ( 5\,{\frac{a}{b}} \right ){\it Si} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) a-5\,{\it Ci} \left ( 5\,\arcsin \left ( dx+c \right ) +5\,{\frac{a}{b}} \right ) \sin \left ( 5\,{\frac{a}{b}} \right ) a+2\,{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-2\,{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a-9\,{\it Si} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a+9\,{\it Ci} \left ( 3\,\arcsin \left ( dx+c \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) a+2\,\sqrt{1- \left ( dx+c \right ) ^{2}}b+\cos \left ( 5\,\arcsin \left ( dx+c \right ) \right ) b-3\,\cos \left ( 3\,\arcsin \left ( dx+c \right ) \right ) b \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.5291, size = 1855, normalized size = 7.19 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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