Optimal. Leaf size=282 \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b^2 c^3}-\frac{3 \sin \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (\frac{6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{\sin \left (\frac{8 a}{b}\right ) \text{CosIntegral}\left (\frac{8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac{\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b^2 c^3}+\frac{3 \cos \left (\frac{6 a}{b}\right ) \text{Si}\left (\frac{6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac{\cos \left (\frac{8 a}{b}\right ) \text{Si}\left (\frac{8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{x^2 \left (1-c^2 x^2\right )^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.933367, antiderivative size = 282, normalized size of antiderivative = 1., 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.
[In]
[Out]
Rule 4721
Rule 4723
Rule 4406
Rule 3303
Rule 3299
Rule 3302
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*}
Mathematica [A] time = 1.08187, size = 414, normalized size = 1.47 \[ \frac{\sin \left (\frac{2 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-2 \sin \left (\frac{4 a}{b}\right ) \left (a+b \sin ^{-1}(c x)\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-3 a \sin \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-3 b \sin \left (\frac{6 a}{b}\right ) \sin ^{-1}(c x) \text{CosIntegral}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-a \sin \left (\frac{8 a}{b}\right ) \text{CosIntegral}\left (8 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-b \sin \left (\frac{8 a}{b}\right ) \sin ^{-1}(c x) \text{CosIntegral}\left (8 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-a \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-b \cos \left (\frac{2 a}{b}\right ) \sin ^{-1}(c x) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 a \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 b \cos \left (\frac{4 a}{b}\right ) \sin ^{-1}(c x) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+3 a \cos \left (\frac{6 a}{b}\right ) \text{Si}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+3 b \cos \left (\frac{6 a}{b}\right ) \sin ^{-1}(c x) \text{Si}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+a \cos \left (\frac{8 a}{b}\right ) \text{Si}\left (8 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+b \cos \left (\frac{8 a}{b}\right ) \sin ^{-1}(c x) \text{Si}\left (8 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+16 b c^8 x^8-48 b c^6 x^6+48 b c^4 x^4-16 b c^2 x^2}{16 b^2 c^3 \left (a+b \sin ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.061, size = 478, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [B] time = 1.8185, size = 3322, normalized size = 11.78 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]