Optimal. Leaf size=242 \[ \frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}-\frac {15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^3}+\frac {15 i b c \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 d^3} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.24, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4701, 4655, 4657, 4181, 2279, 2391, 261, 266, 51, 63, 208} \[ \frac {15 i b c \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}-\frac {15 i c \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 d^3}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^3} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 51
Rule 63
Rule 208
Rule 261
Rule 266
Rule 2279
Rule 2391
Rule 4181
Rule 4655
Rule 4657
Rule 4701
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )^3} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\left (5 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {1}{x \left (1-c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )^{5/2}} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (5 b c^3\right ) \int \frac {x}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac {\left (15 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )^{3/2}} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (15 b c^3\right ) \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac {\left (15 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}+\frac {(15 c) \operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{c d^3}-\frac {(15 b c) \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}+\frac {(15 b c) \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 d^3}\\ &=-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^3}+\frac {(15 i b c) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac {(15 i b c) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 d^3}\\ &=-\frac {b c}{12 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {7 b c}{8 d^3 \sqrt {1-c^2 x^2}}-\frac {a+b \sin ^{-1}(c x)}{d^3 x \left (1-c^2 x^2\right )^2}+\frac {5 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {15 c^2 x \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {15 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 d^3}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^3}+\frac {15 i b c \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}-\frac {15 i b c \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 d^3}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] time = 1.55, size = 512, normalized size = 2.12 \[ -\frac {\frac {14 a c^2 x}{c^2 x^2-1}-\frac {4 a c^2 x}{\left (c^2 x^2-1\right )^2}+15 a c \log (1-c x)-15 a c \log (c x+1)+\frac {16 a}{x}-\frac {b c^2 x \sqrt {1-c^2 x^2}}{3 (c x-1)^2}+\frac {b c^2 x \sqrt {1-c^2 x^2}}{3 (c x+1)^2}-\frac {7 b c \sqrt {1-c^2 x^2}}{c x-1}+\frac {7 b c \sqrt {1-c^2 x^2}}{c x+1}+\frac {2 b c \sqrt {1-c^2 x^2}}{3 (c x-1)^2}+\frac {2 b c \sqrt {1-c^2 x^2}}{3 (c x+1)^2}+16 b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-30 i b c \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+30 i b c \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+\frac {7 b c \sin ^{-1}(c x)}{c x-1}+\frac {7 b c \sin ^{-1}(c x)}{c x+1}-\frac {b c \sin ^{-1}(c x)}{(c x-1)^2}+\frac {b c \sin ^{-1}(c x)}{(c x+1)^2}+15 i \pi b c \sin ^{-1}(c x)+\frac {16 b \sin ^{-1}(c x)}{x}-30 b c \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+30 b c \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-15 \pi b c \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+15 \pi b c \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+15 \pi b c \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{16 d^3} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arcsin \left (c x\right ) + a}{c^{6} d^{3} x^{8} - 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} - d^{3} x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{3} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.33, size = 461, normalized size = 1.90 \[ -\frac {c a}{16 d^{3} \left (c x +1\right )^{2}}-\frac {7 c a}{16 d^{3} \left (c x +1\right )}+\frac {15 c a \ln \left (c x +1\right )}{16 d^{3}}-\frac {a}{d^{3} x}+\frac {c a}{16 d^{3} \left (c x -1\right )^{2}}-\frac {7 c a}{16 d^{3} \left (c x -1\right )}-\frac {15 c a \ln \left (c x -1\right )}{16 d^{3}}-\frac {15 b \arcsin \left (c x \right ) c^{4} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {7 b \,c^{3} x^{2} \sqrt {-c^{2} x^{2}+1}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {25 b \arcsin \left (c x \right ) c^{2} x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {23 c b \sqrt {-c^{2} x^{2}+1}}{24 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b \arcsin \left (c x \right )}{d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right ) x}-\frac {c b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{3}}+\frac {c b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d^{3}}+\frac {15 i c b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {15 i c b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {15 c b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}+\frac {15 c b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{16} \, a {\left (\frac {2 \, {\left (15 \, c^{4} x^{4} - 25 \, c^{2} x^{2} + 8\right )}}{c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x} - \frac {15 \, c \log \left (c x + 1\right )}{d^{3}} + \frac {15 \, c \log \left (c x - 1\right )}{d^{3}}\right )} + \frac {{\left (15 \, {\left (c^{5} x^{5} - 2 \, c^{3} x^{3} + c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) - 15 \, {\left (c^{5} x^{5} - 2 \, c^{3} x^{3} + c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \, {\left (15 \, c^{4} x^{4} - 25 \, c^{2} x^{2} + 8\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - {\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )} \int \frac {{\left (30 \, c^{5} x^{4} - 50 \, c^{3} x^{2} - 15 \, {\left (c^{6} x^{5} - 2 \, c^{4} x^{3} + c^{2} x\right )} \log \left (c x + 1\right ) + 15 \, {\left (c^{6} x^{5} - 2 \, c^{4} x^{3} + c^{2} x\right )} \log \left (-c x + 1\right ) + 16 \, c\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{6} d^{3} x^{7} - 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} - d^{3} x}\,{d x}\right )} b}{16 \, {\left (c^{4} d^{3} x^{5} - 2 \, c^{2} d^{3} x^{3} + d^{3} x\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{6} x^{8} - 3 c^{4} x^{6} + 3 c^{2} x^{4} - x^{2}}\, dx}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________