Integrand size = 15, antiderivative size = 155 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {1215 i \sqrt {3 i x+4 x^2}}{32768}-\frac {135 x \sqrt {3 i x+4 x^2}}{4096}-\frac {9}{256} i x^2 \sqrt {3 i x+4 x^2}-\frac {81}{32} x^3 \sqrt {3 i x+4 x^2}+5 i x^4 \sqrt {3 i x+4 x^2}+\frac {8}{3} x^5 \sqrt {3 i x+4 x^2}+\frac {3645 \text {arctanh}\left (\frac {2 x}{\sqrt {3 i x+4 x^2}}\right )}{65536} \] Output:
1215/32768*I*(3*I*x+4*x^2)^(1/2)-135/4096*x*(3*I*x+4*x^2)^(1/2)-9/256*I*x^ 2*(3*I*x+4*x^2)^(1/2)-81/32*x^3*(3*I*x+4*x^2)^(1/2)+5*I*x^4*(3*I*x+4*x^2)^ (1/2)+8/3*x^5*(3*I*x+4*x^2)^(1/2)+3645/65536*arctanh(2*x/(3*I*x+4*x^2)^(1/ 2))
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {\sqrt {x (3 i+4 x)} \left (7290 i-6480 x-6912 i x^2-497664 x^3+983040 i x^4+524288 x^5-\frac {10935 \log \left (-2 \sqrt {x}+\sqrt {3 i+4 x}\right )}{\sqrt {x} \sqrt {3 i+4 x}}\right )}{196608} \] Input:
Integrate[((3*I)*x + 4*x^2)^(5/2),x]
Output:
(Sqrt[x*(3*I + 4*x)]*(7290*I - 6480*x - (6912*I)*x^2 - 497664*x^3 + (98304 0*I)*x^4 + 524288*x^5 - (10935*Log[-2*Sqrt[x] + Sqrt[3*I + 4*x]])/(Sqrt[x] *Sqrt[3*I + 4*x])))/196608
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.68, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1087, 1087, 1087, 1090, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (4 x^2+3 i x\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {15}{32} \int \left (4 x^2+3 i x\right )^{3/2}dx+\frac {1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {15}{32} \left (\frac {27}{64} \int \sqrt {4 x^2+3 i x}dx+\frac {1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}\right )+\frac {1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {15}{32} \left (\frac {27}{64} \left (\frac {9}{32} \int \frac {1}{\sqrt {4 x^2+3 i x}}dx+\frac {1}{16} \sqrt {4 x^2+3 i x} (8 x+3 i)\right )+\frac {1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}\right )+\frac {1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {15}{32} \left (\frac {27}{64} \left (\frac {3}{64} \int \frac {1}{\sqrt {\frac {1}{9} (8 x+3 i)^2+1}}d(8 x+3 i)+\frac {1}{16} \sqrt {4 x^2+3 i x} (8 x+3 i)\right )+\frac {1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}\right )+\frac {1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {15}{32} \left (\frac {27}{64} \left (\frac {9}{64} \text {arcsinh}\left (\frac {1}{3} (8 x+3 i)\right )+\frac {1}{16} \sqrt {4 x^2+3 i x} (8 x+3 i)\right )+\frac {1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}\right )+\frac {1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}\) |
Input:
Int[((3*I)*x + 4*x^2)^(5/2),x]
Output:
((3*I + 8*x)*((3*I)*x + 4*x^2)^(5/2))/48 + (15*(((3*I + 8*x)*((3*I)*x + 4* x^2)^(3/2))/32 + (27*(((3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/16 + (9*ArcSinh[ (3*I + 8*x)/3])/64))/64))/32
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Time = 0.51 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.37
method | result | size |
risch | \(\frac {\left (262144 x^{5}+491520 i x^{4}-248832 x^{3}-3456 i x^{2}-3240 x +3645 i\right ) x \left (3 i+4 x \right )}{98304 \sqrt {x \left (3 i+4 x \right )}}+\frac {3645 \,\operatorname {arcsinh}\left (i+\frac {8 x}{3}\right )}{131072}\) | \(58\) |
default | \(\frac {\left (3 i+8 x \right ) \left (4 x^{2}+3 i x \right )^{\frac {5}{2}}}{48}+\frac {15 \left (4 x^{2}+3 i x \right )^{\frac {3}{2}} \left (3 i+8 x \right )}{1024}+\frac {405 \left (3 i+8 x \right ) \sqrt {4 x^{2}+3 i x}}{32768}+\frac {3645 \,\operatorname {arcsinh}\left (i+\frac {8 x}{3}\right )}{131072}\) | \(71\) |
trager | \(\left (5 i x^{4}+\frac {8}{3} x^{5}-\frac {9}{256} i x^{2}-\frac {81}{32} x^{3}+\frac {1215}{32768} i-\frac {135}{4096} x \right ) \sqrt {4 x^{2}+3 i x}+\frac {3645 \ln \left (440 x +144+165 i-192 i \sqrt {4 x^{2}+3 i x}-384 i x +220 \sqrt {4 x^{2}+3 i x}\right )}{131072}\) | \(86\) |
Input:
int((3*I*x+4*x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/98304*(3645*I-3240*x-3456*I*x^2-248832*x^3+491520*I*x^4+262144*x^5)*x*(3 *I+4*x)/(x*(3*I+4*x))^(1/2)+3645/131072*arcsinh(I+8/3*x)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.38 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {1}{98304} \, {\left (262144 \, x^{5} + 491520 i \, x^{4} - 248832 \, x^{3} - 3456 i \, x^{2} - 3240 \, x + 3645 i\right )} \sqrt {4 \, x^{2} + 3 i \, x} - \frac {3645}{131072} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 3 i \, x} - \frac {3}{4} i\right ) - \frac {8991}{1048576} \] Input:
integrate((3*I*x+4*x^2)^(5/2),x, algorithm="fricas")
Output:
1/98304*(262144*x^5 + 491520*I*x^4 - 248832*x^3 - 3456*I*x^2 - 3240*x + 36 45*I)*sqrt(4*x^2 + 3*I*x) - 3645/131072*log(-2*x + sqrt(4*x^2 + 3*I*x) - 3 /4*I) - 8991/1048576
Time = 0.61 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=- 9 \sqrt {4 x^{2} + 3 i x} \left (\frac {x^{3}}{4} + \frac {i x^{2}}{32} + \frac {15 x}{512} - \frac {135 i}{4096}\right ) + 16 \sqrt {4 x^{2} + 3 i x} \left (\frac {x^{5}}{6} + \frac {i x^{4}}{80} + \frac {27 x^{3}}{2560} - \frac {189 i x^{2}}{20480} - \frac {567 x}{65536} + \frac {5103 i}{524288}\right ) + 24 i \left (\sqrt {4 x^{2} + 3 i x} \left (\frac {x^{4}}{5} + \frac {3 i x^{3}}{160} + \frac {21 x^{2}}{1280} - \frac {63 i x}{4096} - \frac {567}{32768}\right ) + \frac {1701 i \operatorname {asinh}{\left (\frac {8 x}{3} + i \right )}}{131072}\right ) + \frac {44469 \operatorname {asinh}{\left (\frac {8 x}{3} + i \right )}}{131072} \] Input:
integrate((3*I*x+4*x**2)**(5/2),x)
Output:
-9*sqrt(4*x**2 + 3*I*x)*(x**3/4 + I*x**2/32 + 15*x/512 - 135*I/4096) + 16* sqrt(4*x**2 + 3*I*x)*(x**5/6 + I*x**4/80 + 27*x**3/2560 - 189*I*x**2/20480 - 567*x/65536 + 5103*I/524288) + 24*I*(sqrt(4*x**2 + 3*I*x)*(x**4/5 + 3*I *x**3/160 + 21*x**2/1280 - 63*I*x/4096 - 567/32768) + 1701*I*asinh(8*x/3 + I)/131072) + 44469*asinh(8*x/3 + I)/131072
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (4 \, x^{2} + 3 i \, x\right )}^{\frac {5}{2}} x + \frac {1}{16} i \, {\left (4 \, x^{2} + 3 i \, x\right )}^{\frac {5}{2}} + \frac {15}{128} \, {\left (4 \, x^{2} + 3 i \, x\right )}^{\frac {3}{2}} x + \frac {45}{1024} i \, {\left (4 \, x^{2} + 3 i \, x\right )}^{\frac {3}{2}} + \frac {405}{4096} \, \sqrt {4 \, x^{2} + 3 i \, x} x + \frac {1215}{32768} i \, \sqrt {4 \, x^{2} + 3 i \, x} + \frac {3645}{131072} \, \log \left (8 \, x + 4 \, \sqrt {4 \, x^{2} + 3 i \, x} + 3 i\right ) \] Input:
integrate((3*I*x+4*x^2)^(5/2),x, algorithm="maxima")
Output:
1/6*(4*x^2 + 3*I*x)^(5/2)*x + 1/16*I*(4*x^2 + 3*I*x)^(5/2) + 15/128*(4*x^2 + 3*I*x)^(3/2)*x + 45/1024*I*(4*x^2 + 3*I*x)^(3/2) + 405/4096*sqrt(4*x^2 + 3*I*x)*x + 1215/32768*I*sqrt(4*x^2 + 3*I*x) + 3645/131072*log(8*x + 4*sq rt(4*x^2 + 3*I*x) + 3*I)
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {1}{196608} \, {\left (8 \, {\left (16 \, {\left (8 \, {\left (32 \, {\left (8 \, x + 15 i\right )} x - 243\right )} x - 27 i\right )} x - 405\right )} x + 3645 i\right )} \sqrt {8 \, x^{2} + 2 \, \sqrt {16 \, x^{2} + 9} x} {\left (\frac {3 i \, x}{4 \, x^{2} + \sqrt {16 \, x^{4} + 9 \, x^{2}}} + 1\right )} - \frac {3645}{131072} \, \log \left (2 \, \sqrt {8 \, x^{2} + 2 \, \sqrt {16 \, x^{2} + 9} x} {\left (\frac {3 i \, x}{4 \, x^{2} + \sqrt {16 \, x^{4} + 9 \, x^{2}}} + 1\right )} - 8 \, x - 3 i\right ) \] Input:
integrate((3*I*x+4*x^2)^(5/2),x, algorithm="giac")
Output:
1/196608*(8*(16*(8*(32*(8*x + 15*I)*x - 243)*x - 27*I)*x - 405)*x + 3645*I )*sqrt(8*x^2 + 2*sqrt(16*x^2 + 9)*x)*(3*I*x/(4*x^2 + sqrt(16*x^4 + 9*x^2)) + 1) - 3645/131072*log(2*sqrt(8*x^2 + 2*sqrt(16*x^2 + 9)*x)*(3*I*x/(4*x^2 + sqrt(16*x^4 + 9*x^2)) + 1) - 8*x - 3*I)
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.52 \[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\frac {3645\,\ln \left (x+\frac {\sqrt {x\,\left (4\,x+3{}\mathrm {i}\right )}}{2}+\frac {3}{8}{}\mathrm {i}\right )}{131072}+\frac {15\,\left (4\,x+\frac {3}{2}{}\mathrm {i}\right )\,{\left (4\,x^2+x\,3{}\mathrm {i}\right )}^{3/2}}{512}+\frac {\left (4\,x+\frac {3}{2}{}\mathrm {i}\right )\,{\left (4\,x^2+x\,3{}\mathrm {i}\right )}^{5/2}}{24}+\frac {405\,\left (\frac {x}{2}+\frac {3}{16}{}\mathrm {i}\right )\,\sqrt {4\,x^2+x\,3{}\mathrm {i}}}{2048} \] Input:
int((x*3i + 4*x^2)^(5/2),x)
Output:
(3645*log(x + (x*(4*x + 3i))^(1/2)/2 + 3i/8))/131072 + (15*(4*x + 3i/2)*(x *3i + 4*x^2)^(3/2))/512 + ((4*x + 3i/2)*(x*3i + 4*x^2)^(5/2))/24 + (405*(x /2 + 3i/16)*(x*3i + 4*x^2)^(1/2))/2048
\[ \int \left (3 i x+4 x^2\right )^{5/2} \, dx=\int \left (4 x^{2}+3 i x \right )^{\frac {5}{2}}d x \] Input:
int((3*I*x+4*x^2)^(5/2),x)
Output:
int((3*I*x+4*x^2)^(5/2),x)