Optimal. Leaf size=282 \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b^2 c^3}-\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {Ci}\left (\frac {8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]
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Rubi [A] time = 0.93, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4721, 4723, 4406, 3303, 3299, 3302} \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b^2 c^3}-\frac {3 \sin \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {CosIntegral}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{16 b^2 c^3}-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
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Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rule 4721
Rule 4723
Rubi steps
\begin {align*} \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \int \frac {x \left (1-c^2 x^2\right )^2}{a+b \sin ^{-1}(c x)} \, dx}{b c}-\frac {(8 c) \int \frac {x^3 \left (1-c^2 x^2\right )^2}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {\cos ^5(x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {8 \operatorname {Subst}\left (\int \frac {\cos ^5(x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {5 \sin (2 x)}{32 (a+b x)}+\frac {\sin (4 x)}{8 (a+b x)}+\frac {\sin (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {8 \operatorname {Subst}\left (\int \left (\frac {3 \sin (2 x)}{64 (a+b x)}+\frac {\sin (4 x)}{64 (a+b x)}-\frac {\sin (6 x)}{64 (a+b x)}-\frac {\sin (8 x)}{128 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sin (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\operatorname {Subst}\left (\int \frac {\sin (8 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\operatorname {Subst}\left (\int \frac {\sin (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {5 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\left (5 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\left (3 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\cos \left (\frac {8 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {8 a}{b}+8 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\left (5 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\left (3 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\sin \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {8 a}{b}+8 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c^3}-\frac {\text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{8 b^2 c^3}-\frac {3 \text {Ci}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c^3}-\frac {\text {Ci}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right ) \sin \left (\frac {8 a}{b}\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{16 b^2 c^3}\\ \end {align*}
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Mathematica [A] time = 1.22, size = 414, normalized size = 1.47 \[ \frac {\sin \left (\frac {2 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-2 \sin \left (\frac {4 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 a \sin \left (\frac {6 a}{b}\right ) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-3 b \sin \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-a \sin \left (\frac {8 a}{b}\right ) \text {Ci}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-b \sin \left (\frac {8 a}{b}\right ) \sin ^{-1}(c x) \text {Ci}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-b \cos \left (\frac {2 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+2 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+2 b \cos \left (\frac {4 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+3 b \cos \left (\frac {6 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+a \cos \left (\frac {8 a}{b}\right ) \text {Si}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+b \cos \left (\frac {8 a}{b}\right ) \sin ^{-1}(c x) \text {Si}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+16 b c^8 x^8-48 b c^6 x^6+48 b c^4 x^4-16 b c^2 x^2}{16 b^2 c^3 \left (a+b \sin ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{6} - 2 \, c^{2} x^{4} + x^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.69, size = 2461, normalized size = 8.73 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 478, normalized size = 1.70 \[ \frac {16 \arcsin \left (c x \right ) \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -16 \arcsin \left (c x \right ) \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +8 \arcsin \left (c x \right ) \Si \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \cos \left (\frac {8 a}{b}\right ) b -8 \arcsin \left (c x \right ) \Ci \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \sin \left (\frac {8 a}{b}\right ) b -8 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +8 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +24 \arcsin \left (c x \right ) \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b -24 \arcsin \left (c x \right ) \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b +16 \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -16 \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +8 \Si \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \cos \left (\frac {8 a}{b}\right ) a -8 \Ci \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \sin \left (\frac {8 a}{b}\right ) a -8 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +8 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +24 \Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a -24 \Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a +4 \cos \left (4 \arcsin \left (c x \right )\right ) b +\cos \left (8 \arcsin \left (c x \right )\right ) b -4 \cos \left (2 \arcsin \left (c x \right )\right ) b +4 \cos \left (6 \arcsin \left (c x \right )\right ) b -5 b}{128 c^{3} \left (a +b \arcsin \left (c x \right )\right ) b^{2}} \]
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 \[ \frac {c^{6} x^{8} - 3 \, c^{4} x^{6} + 3 \, c^{2} x^{4} - x^{2} - 2 \, {\left (b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c\right )} \int \frac {4 \, c^{6} x^{7} - 9 \, c^{4} x^{5} + 6 \, c^{2} x^{3} - x}{b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c}\,{d x}}{b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{5/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \]
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 \[ \int \frac {x^{2} \left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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