\(\int \sqrt {c+d x} \cosh (a+b x) \, dx\) [43]

Optimal result
Mathematica [A] (verified)
Rubi [C] (verified)
Maple [F]
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 123 \[ \int \sqrt {c+d x} \cosh (a+b x) \, dx=\frac {\sqrt {d} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}-\frac {\sqrt {d} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}+\frac {\sqrt {c+d x} \sinh (a+b x)}{b} \] Output:

1/4*d^(1/2)*exp(-a+b*c/d)*Pi^(1/2)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/b^(3 
/2)-1/4*d^(1/2)*exp(a-b*c/d)*Pi^(1/2)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/ 
b^(3/2)+(d*x+c)^(1/2)*sinh(b*x+a)/b
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.85 \[ \int \sqrt {c+d x} \cosh (a+b x) \, dx=\frac {e^{-a-\frac {b c}{d}} \sqrt {c+d x} \left (\frac {e^{2 a} \Gamma \left (\frac {3}{2},-\frac {b (c+d x)}{d}\right )}{\sqrt {-\frac {b (c+d x)}{d}}}-\frac {e^{\frac {2 b c}{d}} \Gamma \left (\frac {3}{2},\frac {b (c+d x)}{d}\right )}{\sqrt {\frac {b (c+d x)}{d}}}\right )}{2 b} \] Input:

Integrate[Sqrt[c + d*x]*Cosh[a + b*x],x]
 

Output:

(E^(-a - (b*c)/d)*Sqrt[c + d*x]*((E^(2*a)*Gamma[3/2, -((b*(c + d*x))/d)])/ 
Sqrt[-((b*(c + d*x))/d)] - (E^((2*b*c)/d)*Gamma[3/2, (b*(c + d*x))/d])/Sqr 
t[(b*(c + d*x))/d]))/(2*b)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 3777, 26, 3042, 26, 3789, 2611, 2633, 2634}

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.

\(\displaystyle \int \sqrt {c+d x} \cosh (a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {c+d x} \sin \left (i a+i b x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}-\frac {i d \int -\frac {i \sinh (a+b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}-\frac {d \int \frac {\sinh (a+b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}-\frac {d \int -\frac {i \sin (i a+i b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}+\frac {i d \int \frac {\sin (i a+i b x)}{\sqrt {c+d x}}dx}{2 b}\)

\(\Big \downarrow \) 3789

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}+\frac {i d \left (\frac {1}{2} i \int \frac {e^{a+b x}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {e^{-a-b x}}{\sqrt {c+d x}}dx\right )}{2 b}\)

\(\Big \downarrow \) 2611

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}+\frac {i d \left (\frac {i \int e^{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{2 b}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}+\frac {i d \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{2 b}\)

\(\Big \downarrow \) 2634

\(\displaystyle \frac {\sqrt {c+d x} \sinh (a+b x)}{b}+\frac {i d \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{2 b}\)

Input:

Int[Sqrt[c + d*x]*Cosh[a + b*x],x]
 

Output:

((I/2)*d*(((-1/2*I)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/ 
Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + ((I/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b 
]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d])))/b + (Sqrt[c + d*x]*Sinh[a + 
 b*x])/b
 

Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2611
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2633
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2634
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3777
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3789
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
/2   Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2   Int[(c + d*x)^m*E 
^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
 
Maple [F]

\[\int \sqrt {d x +c}\, \cosh \left (b x +a \right )d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (91) = 182\).

Time = 0.11 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.46 \[ \int \sqrt {c+d x} \cosh (a+b x) \, dx=\frac {\sqrt {\pi } {\left (d \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - d \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d \cosh \left (-\frac {b c - a d}{d}\right ) - d \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) + \sqrt {\pi } {\left (d \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + d \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d \cosh \left (-\frac {b c - a d}{d}\right ) + d \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) + 2 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )} \sqrt {d x + c}}{4 \, {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \] Input:

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

Output:

1/4*(sqrt(pi)*(d*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - d*cosh(b*x + a)*sinh 
(-(b*c - a*d)/d) + (d*cosh(-(b*c - a*d)/d) - d*sinh(-(b*c - a*d)/d))*sinh( 
b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) + sqrt(pi)*(d*cosh(b*x + 
a)*cosh(-(b*c - a*d)/d) + d*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + (d*cosh(- 
(b*c - a*d)/d) + d*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqr 
t(d*x + c)*sqrt(-b/d)) + 2*(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x 
 + a) + b*sinh(b*x + a)^2 - b)*sqrt(d*x + c))/(b^2*cosh(b*x + a) + b^2*sin 
h(b*x + a))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (91) = 182\).

Time = 0.05 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.87 \[ \int \sqrt {c+d x} \cosh (a+b x) \, dx=\frac {8 \, {\left (d x + c\right )}^{\frac {3}{2}} \cosh \left (b x + a\right ) - \frac {{\left (\frac {3 \, \sqrt {\pi } d^{2} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{b^{2} \sqrt {-\frac {b}{d}}} - \frac {3 \, \sqrt {\pi } d^{2} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{b^{2} \sqrt {\frac {b}{d}}} + \frac {2 \, {\left (2 \, {\left (d x + c\right )}^{\frac {3}{2}} b d e^{\left (\frac {b c}{d}\right )} + 3 \, \sqrt {d x + c} d^{2} e^{\left (\frac {b c}{d}\right )}\right )} e^{\left (-a - \frac {{\left (d x + c\right )} b}{d}\right )}}{b^{2}} + \frac {2 \, {\left (2 \, {\left (d x + c\right )}^{\frac {3}{2}} b d e^{a} - 3 \, \sqrt {d x + c} d^{2} e^{a}\right )} e^{\left (\frac {{\left (d x + c\right )} b}{d} - \frac {b c}{d}\right )}}{b^{2}}\right )} b}{d}}{12 \, d} \] Input:

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

Output:

1/12*(8*(d*x + c)^(3/2)*cosh(b*x + a) - (3*sqrt(pi)*d^2*erf(sqrt(d*x + c)* 
sqrt(-b/d))*e^(a - b*c/d)/(b^2*sqrt(-b/d)) - 3*sqrt(pi)*d^2*erf(sqrt(d*x + 
 c)*sqrt(b/d))*e^(-a + b*c/d)/(b^2*sqrt(b/d)) + 2*(2*(d*x + c)^(3/2)*b*d*e 
^(b*c/d) + 3*sqrt(d*x + c)*d^2*e^(b*c/d))*e^(-a - (d*x + c)*b/d)/b^2 + 2*( 
2*(d*x + c)^(3/2)*b*d*e^a - 3*sqrt(d*x + c)*d^2*e^a)*e^((d*x + c)*b/d - b* 
c/d)/b^2)*b/d)/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.37 \[ \int \sqrt {c+d x} \cosh (a+b x) \, dx=-\frac {\frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c}}{d}\right ) e^{\left (\frac {b c - a d}{d}\right )}}{\sqrt {b d} b} - \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {\sqrt {-b d} \sqrt {d x + c}}{d}\right ) e^{\left (-\frac {b c - a d}{d}\right )}}{\sqrt {-b d} b} - \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b} + \frac {2 \, \sqrt {d x + c} d e^{\left (-\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b}}{4 \, d} \] Input:

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

Output:

-1/4*(sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)/d)*e^((b*c - a*d)/d)/(sqrt 
(b*d)*b) - sqrt(pi)*d^2*erf(-sqrt(-b*d)*sqrt(d*x + c)/d)*e^(-(b*c - a*d)/d 
)/(sqrt(-b*d)*b) - 2*sqrt(d*x + c)*d*e^(((d*x + c)*b - b*c + a*d)/d)/b + 2 
*sqrt(d*x + c)*d*e^(-((d*x + c)*b - b*c + a*d)/d)/b)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(c + d*x)*cosh(a + b*x),x)