\(\int x^3 (a+b \tan (c+d \sqrt {x})) \, dx\) [25]

Optimal result

Integrand size = 18, antiderivative size = 261 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \] Output:

1/4*a*x^4+1/4*I*b*x^4-2*b*x^(7/2)*ln(1+exp(2*I*(c+d*x^(1/2))))/d+7*I*b*x^3 
*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^2-21*b*x^(5/2)*polylog(3,-exp(2*I*(c 
+d*x^(1/2))))/d^3-105/2*I*b*x^2*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^4+105 
*b*x^(3/2)*polylog(5,-exp(2*I*(c+d*x^(1/2))))/d^5+315/2*I*b*x*polylog(6,-e 
xp(2*I*(c+d*x^(1/2))))/d^6-315/2*b*x^(1/2)*polylog(7,-exp(2*I*(c+d*x^(1/2) 
)))/d^7-315/4*I*b*polylog(8,-exp(2*I*(c+d*x^(1/2))))/d^8
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \] Input:

Integrate[x^3*(a + b*Tan[c + d*Sqrt[x]]),x]
 

Output:

(a*x^4)/4 + (I/4)*b*x^4 - (2*b*x^(7/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))]) 
/d + ((7*I)*b*x^3*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 - (21*b*x^(5 
/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (((105*I)/2)*b*x^2*PolyL 
og[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 + (105*b*x^(3/2)*PolyLog[5, -E^((2* 
I)*(c + d*Sqrt[x]))])/d^5 + (((315*I)/2)*b*x*PolyLog[6, -E^((2*I)*(c + d*S 
qrt[x]))])/d^6 - (315*b*Sqrt[x]*PolyLog[7, -E^((2*I)*(c + d*Sqrt[x]))])/(2 
*d^7) - (((315*I)/4)*b*PolyLog[8, -E^((2*I)*(c + d*Sqrt[x]))])/d^8
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2010, 2009}

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.

\(\displaystyle \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (a x^3+b x^3 \tan \left (c+d \sqrt {x}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^4}{4}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{4} i b x^4\)

Input:

Int[x^3*(a + b*Tan[c + d*Sqrt[x]]),x]
 

Output:

(a*x^4)/4 + (I/4)*b*x^4 - (2*b*x^(7/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))]) 
/d + ((7*I)*b*x^3*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 - (21*b*x^(5 
/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (((105*I)/2)*b*x^2*PolyL 
og[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 + (105*b*x^(3/2)*PolyLog[5, -E^((2* 
I)*(c + d*Sqrt[x]))])/d^5 + (((315*I)/2)*b*x*PolyLog[6, -E^((2*I)*(c + d*S 
qrt[x]))])/d^6 - (315*b*Sqrt[x]*PolyLog[7, -E^((2*I)*(c + d*Sqrt[x]))])/(2 
*d^7) - (((315*I)/4)*b*PolyLog[8, -E^((2*I)*(c + d*Sqrt[x]))])/d^8
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

\[\int x^{3} \left (a +b \tan \left (c +d \sqrt {x}\right )\right )d x\]

Input:

int(x^3*(a+b*tan(c+d*x^(1/2))),x)
 

Output:

int(x^3*(a+b*tan(c+d*x^(1/2))),x)
 

Fricas [F]

\[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*tan(c+d*x^(1/2))),x, algorithm="fricas")
 

Output:

integral(b*x^3*tan(d*sqrt(x) + c) + a*x^3, x)
 

Sympy [F]

\[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int x^{3} \left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )\, dx \] Input:

integrate(x**3*(a+b*tan(c+d*x**(1/2))),x)
 

Output:

Integral(x**3*(a + b*tan(c + d*sqrt(x))), x)
 

Maxima [B] (verification not implemented)

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 = 0.20 (sec) , antiderivative size = 937, normalized size of antiderivative = 3.59 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\text {Too large to display} \] Input:

integrate(x^3*(a+b*tan(c+d*x^(1/2))),x, algorithm="maxima")
 

Output:

1/420*(105*(d*sqrt(x) + c)^8*a + 105*I*(d*sqrt(x) + c)^8*b - 840*(d*sqrt(x 
) + c)^7*a*c - 840*I*(d*sqrt(x) + c)^7*b*c + 2940*(d*sqrt(x) + c)^6*a*c^2 
+ 2940*I*(d*sqrt(x) + c)^6*b*c^2 - 5880*(d*sqrt(x) + c)^5*a*c^3 - 5880*I*( 
d*sqrt(x) + c)^5*b*c^3 + 7350*(d*sqrt(x) + c)^4*a*c^4 + 7350*I*(d*sqrt(x) 
+ c)^4*b*c^4 - 5880*(d*sqrt(x) + c)^3*a*c^5 - 5880*I*(d*sqrt(x) + c)^3*b*c 
^5 + 2940*(d*sqrt(x) + c)^2*a*c^6 + 2940*I*(d*sqrt(x) + c)^2*b*c^6 - 840*( 
d*sqrt(x) + c)*a*c^7 - 840*b*c^7*log(sec(d*sqrt(x) + c)) + 8*(-960*I*(d*sq 
rt(x) + c)^7*b + 3920*I*(d*sqrt(x) + c)^6*b*c - 7056*I*(d*sqrt(x) + c)^5*b 
*c^2 + 7350*I*(d*sqrt(x) + c)^4*b*c^3 - 4900*I*(d*sqrt(x) + c)^3*b*c^4 + 2 
205*I*(d*sqrt(x) + c)^2*b*c^5 - 735*I*(d*sqrt(x) + c)*b*c^6)*arctan2(sin(2 
*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) + 420*(64*I*(d*sqrt(x) + c) 
^6*b - 224*I*(d*sqrt(x) + c)^5*b*c + 336*I*(d*sqrt(x) + c)^4*b*c^2 - 280*I 
*(d*sqrt(x) + c)^3*b*c^3 + 140*I*(d*sqrt(x) + c)^2*b*c^4 - 42*I*(d*sqrt(x) 
 + c)*b*c^5 + 7*I*b*c^6)*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - 4*(960*(d*sqr 
t(x) + c)^7*b - 3920*(d*sqrt(x) + c)^6*b*c + 7056*(d*sqrt(x) + c)^5*b*c^2 
- 7350*(d*sqrt(x) + c)^4*b*c^3 + 4900*(d*sqrt(x) + c)^3*b*c^4 - 2205*(d*sq 
rt(x) + c)^2*b*c^5 + 735*(d*sqrt(x) + c)*b*c^6)*log(cos(2*d*sqrt(x) + 2*c) 
^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos(2*d*sqrt(x) + 2*c) + 1) - 302400*I*b 
*polylog(8, -e^(2*I*d*sqrt(x) + 2*I*c)) - 50400*(12*(d*sqrt(x) + c)*b - 7* 
b*c)*polylog(7, -e^(2*I*d*sqrt(x) + 2*I*c)) + 10080*(60*I*(d*sqrt(x) + ...
 

Giac [F]

\[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int { {\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*tan(c+d*x^(1/2))),x, algorithm="giac")
 

Output:

integrate((b*tan(d*sqrt(x) + c) + a)*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\int x^3\,\left (a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )\right ) \,d x \] Input:

int(x^3*(a + b*tan(c + d*x^(1/2))),x)
 

Output:

int(x^3*(a + b*tan(c + d*x^(1/2))), x)
 

Reduce [F]

\[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\left (\int \tan \left (\sqrt {x}\, d +c \right ) x^{3}d x \right ) b +\frac {a \,x^{4}}{4} \] Input:

int(x^3*(a+b*tan(c+d*x^(1/2))),x)
 

Output:

(4*int(tan(sqrt(x)*d + c)*x**3,x)*b + a*x**4)/4