Optimal. Leaf size=261 \[ \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \text {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \text {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \]
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Rubi [A]
time = 0.26, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {14, 3832,
3800, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {a x^4}{4}-\frac {315 i b \text {Li}_8\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}-\frac {315 b \sqrt {x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}+\frac {315 i b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{4} i b x^4 \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2221
Rule 2320
Rule 2611
Rule 3800
Rule 3832
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \tan \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \tan \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^4}{4}+(2 b) \text {Subst}\left (\int x^7 \tan (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-(4 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^7}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {(14 b) \text {Subst}\left (\int x^6 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(42 i b) \text {Subst}\left (\int x^5 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(105 b) \text {Subst}\left (\int x^4 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {(210 i b) \text {Subst}\left (\int x^3 \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(315 b) \text {Subst}\left (\int x^2 \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {(315 i b) \text {Subst}\left (\int x \text {Li}_6\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}+\frac {(315 b) \text {Subst}\left (\int \text {Li}_7\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{2 d^7}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {(315 i b) \text {Subst}\left (\int \frac {\text {Li}_7(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}\\ &=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \text {Li}_8\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}\\ \end {align*}
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Mathematica [A]
time = 0.10, size = 261, normalized size = 1.00 \begin {gather*} \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \text {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \text {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \text {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \text {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int x^{3} \left (a +b \tan \left (c +d \sqrt {x}\right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 937 vs. \(2 (198) = 396\).
time = 0.57, size = 937, normalized size = 3.59 \begin {gather*} \frac {105 \, {\left (d \sqrt {x} + c\right )}^{8} a + 105 i \, {\left (d \sqrt {x} + c\right )}^{8} b - 840 \, {\left (d \sqrt {x} + c\right )}^{7} a c - 840 i \, {\left (d \sqrt {x} + c\right )}^{7} b c + 2940 \, {\left (d \sqrt {x} + c\right )}^{6} a c^{2} + 2940 i \, {\left (d \sqrt {x} + c\right )}^{6} b c^{2} - 5880 \, {\left (d \sqrt {x} + c\right )}^{5} a c^{3} - 5880 i \, {\left (d \sqrt {x} + c\right )}^{5} b c^{3} + 7350 \, {\left (d \sqrt {x} + c\right )}^{4} a c^{4} + 7350 i \, {\left (d \sqrt {x} + c\right )}^{4} b c^{4} - 5880 \, {\left (d \sqrt {x} + c\right )}^{3} a c^{5} - 5880 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{5} + 2940 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{6} + 2940 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{6} - 840 \, {\left (d \sqrt {x} + c\right )} a c^{7} - 840 \, b c^{7} \log \left (\sec \left (d \sqrt {x} + c\right )\right ) + 8 \, {\left (-960 i \, {\left (d \sqrt {x} + c\right )}^{7} b + 3920 i \, {\left (d \sqrt {x} + c\right )}^{6} b c - 7056 i \, {\left (d \sqrt {x} + c\right )}^{5} b c^{2} + 7350 i \, {\left (d \sqrt {x} + c\right )}^{4} b c^{3} - 4900 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{4} + 2205 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{5} - 735 i \, {\left (d \sqrt {x} + c\right )} b c^{6}\right )} \arctan \left (\sin \left (2 \, d \sqrt {x} + 2 \, c\right ), \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) + 420 \, {\left (64 i \, {\left (d \sqrt {x} + c\right )}^{6} b - 224 i \, {\left (d \sqrt {x} + c\right )}^{5} b c + 336 i \, {\left (d \sqrt {x} + c\right )}^{4} b c^{2} - 280 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{3} + 140 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{4} - 42 i \, {\left (d \sqrt {x} + c\right )} b c^{5} + 7 i \, b c^{6}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}\right ) - 4 \, {\left (960 \, {\left (d \sqrt {x} + c\right )}^{7} b - 3920 \, {\left (d \sqrt {x} + c\right )}^{6} b c + 7056 \, {\left (d \sqrt {x} + c\right )}^{5} b c^{2} - 7350 \, {\left (d \sqrt {x} + c\right )}^{4} b c^{3} + 4900 \, {\left (d \sqrt {x} + c\right )}^{3} b c^{4} - 2205 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{5} + 735 \, {\left (d \sqrt {x} + c\right )} b c^{6}\right )} \log \left (\cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) + 1\right ) - 302400 i \, b {\rm Li}_{8}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) - 50400 \, {\left (12 \, {\left (d \sqrt {x} + c\right )} b - 7 \, b c\right )} {\rm Li}_{7}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) + 10080 \, {\left (60 i \, {\left (d \sqrt {x} + c\right )}^{2} b - 70 i \, {\left (d \sqrt {x} + c\right )} b c + 21 i \, b c^{2}\right )} {\rm Li}_{6}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) + 2520 \, {\left (160 \, {\left (d \sqrt {x} + c\right )}^{3} b - 280 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 168 \, {\left (d \sqrt {x} + c\right )} b c^{2} - 35 \, b c^{3}\right )} {\rm Li}_{5}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) + 840 \, {\left (-240 i \, {\left (d \sqrt {x} + c\right )}^{4} b + 560 i \, {\left (d \sqrt {x} + c\right )}^{3} b c - 504 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} + 210 i \, {\left (d \sqrt {x} + c\right )} b c^{3} - 35 i \, b c^{4}\right )} {\rm Li}_{4}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}) - 420 \, {\left (192 \, {\left (d \sqrt {x} + c\right )}^{5} b - 560 \, {\left (d \sqrt {x} + c\right )}^{4} b c + 672 \, {\left (d \sqrt {x} + c\right )}^{3} b c^{2} - 420 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{3} + 140 \, {\left (d \sqrt {x} + c\right )} b c^{4} - 21 \, b c^{5}\right )} {\rm Li}_{3}(-e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )})}{420 \, d^{8}} \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 x^{3} \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^3\,\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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