Optimal. Leaf size=186 \[ \frac {e^2 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b^2 d}-\frac {3 e^2 \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b^2 d}-\frac {e^2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{4 b^2 d}+\frac {3 e^2 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b^2 d}-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
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Rubi [A] time = 0.26, antiderivative size = 182, normalized size of antiderivative = 0.98, number of steps used = 10, 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} \[ \frac {e^2 \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac {3 e^2 \sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac {e^2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}+\frac {3 e^2 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 3302
Rule 3303
Rule 4631
Rule 4805
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^2 x^2}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^2 \operatorname {Subst}\left (\int \left (-\frac {\sin (x)}{4 (a+b x)}+\frac {3 \sin (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^2 \operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 b d}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 b d}\\ &=-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\left (e^2 \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 )}{4 b d}+\frac {\left (3 e^2 \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 )}{4 b d}+\frac {\left (e^2 \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 )}{4 b d}-\frac {\left (3 e^2 \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 )}{4 b d}\\ &=-\frac {e^2 (c+d x)^2 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^2 \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right ) \sin \left (\frac {a}{b}\right )}{4 b^2 d}-\frac {3 e^2 \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right ) \sin \left (\frac {3 a}{b}\right )}{4 b^2 d}-\frac {e^2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{4 b^2 d}+\frac {3 e^2 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{4 b^2 d}\\ \end {align*}
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Mathematica [A] time = 0.85, size = 140, normalized size = 0.75 \[ \frac {e^2 \left (\sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-3 \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-\frac {4 b \sqrt {1-(c+d x)^2} (c+d x)^2}{a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.39, size = 684, normalized size = 3.68 \[ -\frac {3 \, b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {3 \, b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{3} e^{2} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {3 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {3 \, a \cos \left (\frac {a}{b}\right )^{3} e^{2} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {3 \, b \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} + \frac {b \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {9 \, b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) e^{2} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) e^{2} \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} + \frac {3 \, a \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} + \frac {a \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right ) e^{2} \sin \left (\frac {a}{b}\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {9 \, a \cos \left (\frac {a}{b}\right ) e^{2} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (d x + c\right )\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} - \frac {a \cos \left (\frac {a}{b}\right ) e^{2} \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{4 \, {\left (b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d\right )}} + \frac {{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} b e^{2}}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} b e^{2}}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 266, normalized size = 1.43 \[ -\frac {e^{2} \left (\arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -\arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b -3 \arcsin \left (d x +c \right ) \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b +3 \arcsin \left (d x +c \right ) \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b +\Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -\Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a -3 \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a +3 \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a -\cos \left (3 \arcsin \left (d x +c \right )\right ) b +\sqrt {1-\left (d x +c \right )^{2}}\, b \right )}{4 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^2}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2} \,d x \]
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 \[ e^{2} \left (\int \frac {c^{2}}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{2} x^{2}}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {2 c d x}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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