Optimal. Leaf size=322 \[ -\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {1-(c+d x)^2} (c+d x)^4}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]
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Rubi [A] time = 0.86, antiderivative size = 318, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302} \[ -\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {1-(c+d x)^2} (c+d x)^4}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rule 4633
Rule 4635
Rule 4719
Rule 4805
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^4}{\left (a+b \sin ^{-1}(c+d x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}-\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {x^4}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 (a+b x)}-\frac {\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{8 (a+b x)}-\frac {3 \cos (3 x)}{16 (a+b x)}+\frac {\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {\left (75 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (3 e^4 \cos \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 )}{2 b^2 d}-\frac {\left (25 e^4 \cos \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 )}{16 b^2 d}-\frac {\left (3 e^4 \cos \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 )}{2 b^2 d}+\frac {\left (75 e^4 \cos \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 )}{32 b^2 d}-\frac {\left (25 e^4 \cos \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}+\frac {\left (3 e^4 \sin \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 )}{2 b^2 d}-\frac {\left (25 e^4 \sin \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 )}{16 b^2 d}-\frac {\left (3 e^4 \sin \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 )}{2 b^2 d}+\frac {\left (75 e^4 \sin \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 )}{32 b^2 d}-\frac {\left (25 e^4 \sin \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}\\ \end {align*}
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Mathematica [A] time = 1.60, size = 317, normalized size = 0.98 \[ \frac {e^4 \left (-\frac {16 b^2 \sqrt {1-(c+d x)^2} (c+d x)^4}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+48 \left (\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )-25 \left (2 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )+\frac {16 b \left (5 (c+d x)^5-4 (c+d x)^3\right )}{a+b \sin ^{-1}(c+d x)}\right )}{32 b^3 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.72, size = 3135, normalized size = 9.74 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.26, size = 720, normalized size = 2.24 \[ \frac {e^{4} \left (27 \arcsin \left (d x +c \right )^{2} \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b^{2}-25 \arcsin \left (d x +c \right )^{2} \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b^{2}-25 \arcsin \left (d x +c \right )^{2} \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b^{2}-2 \arcsin \left (d x +c \right )^{2} \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b^{2}-2 \arcsin \left (d x +c \right )^{2} \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b^{2}+27 \arcsin \left (d x +c \right )^{2} \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b^{2}-4 \arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a b -4 \arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a b +54 \arcsin \left (d x +c \right ) \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a b +54 \arcsin \left (d x +c \right ) \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a b -50 \arcsin \left (d x +c \right ) \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a b -9 \sin \left (3 \arcsin \left (d x +c \right )\right ) \arcsin \left (d x +c \right ) b^{2}-2 \sqrt {1-\left (d x +c \right )^{2}}\, b^{2}+3 \cos \left (3 \arcsin \left (d x +c \right )\right ) b^{2}-\cos \left (5 \arcsin \left (d x +c \right )\right ) b^{2}-50 \arcsin \left (d x +c \right ) \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a b -25 \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a^{2}-25 \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a^{2}-2 \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a^{2}-2 \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a^{2}+27 \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a^{2}+27 \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a^{2}+5 \arcsin \left (d x +c \right ) \sin \left (5 \arcsin \left (d x +c \right )\right ) b^{2}+5 \sin \left (5 \arcsin \left (d x +c \right )\right ) a b +2 \arcsin \left (d x +c \right ) \left (d x +c \right ) b^{2}+2 \left (d x +c \right ) a b -9 \sin \left (3 \arcsin \left (d x +c \right )\right ) a b \right )}{32 d \left (a +b \arcsin \left (d x +c \right )\right )^{2} b^{3}} \]
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 = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3} \,d x \]
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 \[ e^{4} \left (\int \frac {c^{4}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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