Optimal. Leaf size=155 \[ -\frac {\left (1-(a+b x)^2\right )^2}{3 b \text {ArcSin}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \text {ArcSin}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \text {ArcSin}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \text {ArcSin}(a+b x)}+\frac {2 \text {Si}(2 \text {ArcSin}(a+b x))}{3 b}+\frac {4 \text {Si}(4 \text {ArcSin}(a+b x))}{3 b} \]
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Rubi [A]
time = 0.24, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps
used = 18, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4891, 4751,
4799, 4731, 4491, 12, 3380} \begin {gather*} \frac {2 \text {Si}(2 \text {ArcSin}(a+b x))}{3 b}+\frac {4 \text {Si}(4 \text {ArcSin}(a+b x))}{3 b}-\frac {8 \left (1-(a+b x)^2\right ) (a+b x)^2}{3 b \text {ArcSin}(a+b x)}+\frac {2 \left (1-(a+b x)^2\right )^{3/2} (a+b x)}{3 b \text {ArcSin}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \text {ArcSin}(a+b x)}-\frac {\left (1-(a+b x)^2\right )^2}{3 b \text {ArcSin}(a+b x)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3380
Rule 4491
Rule 4731
Rule 4751
Rule 4799
Rule 4891
Rubi steps
\begin {align*} \int \frac {\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\sin ^{-1}(a+b x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}-\frac {4 \text {Subst}\left (\int \frac {x \left (1-x^2\right )}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}-\frac {2 \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}+\frac {8 \text {Subst}\left (\int \frac {x^2 \sqrt {1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}+\frac {16 \text {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}-\frac {32 \text {Subst}\left (\int \frac {x^3}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac {16 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac {32 \text {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {4 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac {16 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac {32 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 x}-\frac {\sin (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}+\frac {4 \text {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac {4 \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}\\ \end {align*}
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Mathematica [A]
time = 0.27, size = 143, normalized size = 0.92 \begin {gather*} \frac {\frac {\left (-1+a^2+2 a b x+b^2 x^2\right ) \left (1-a^2-2 a b x-b^2 x^2-2 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)+2 \left (-1+4 a^2+8 a b x+4 b^2 x^2\right ) \text {ArcSin}(a+b x)^2\right )}{\text {ArcSin}(a+b x)^3}+2 \text {Si}(2 \text {ArcSin}(a+b x))+4 \text {Si}(4 \text {ArcSin}(a+b x))}{3 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.42, size = 148, normalized size = 0.95
method | result | size |
default | \(\frac {16 \sinIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+32 \sinIntegral \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+8 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+8 \cos \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+4 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+2 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-4 \cos \left (2 \arcsin \left (b x +a \right )\right )-\cos \left (4 \arcsin \left (b x +a \right )\right )-3}{24 b \arcsin \left (b x +a \right )^{3}}\) | \(148\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\operatorname {asin}^{4}{\left (a + b x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.50, size = 163, normalized size = 1.05 \begin {gather*} \frac {8 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )} + \frac {4 \, \operatorname {Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac {2 \, \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac {2 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{b \arcsin \left (b x + a\right )} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )^{3}} \end {gather*}
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 \begin {gather*} \int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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