\(\int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx\) [48]

Optimal result

Integrand size = 18, antiderivative size = 139 \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=-\frac {\sqrt {c+d x}}{d}+\frac {e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {e^{2 a-\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}} \]

[Out]

1/8*exp(-2*a+2*b*c/d)*erf(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/b^(1/2)/d^(1/2)+1/8*exp(2*a-
2*b*c/d)*erfi(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/b^(1/2)/d^(1/2)-(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3393, 3388, 2211, 2235, 2236} \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=\frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {\sqrt {\frac {\pi }{2}} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}-\frac {\sqrt {c+d x}}{d} \]

[In]

Int[Sinh[a + b*x]^2/Sqrt[c + d*x],x]

[Out]

-(Sqrt[c + d*x]/d) + (E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*Sqrt[b]
*Sqrt[d]) + (E^(2*a - (2*b*c)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*Sqrt[b]*Sqrt[d])

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 2235

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 2236

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps \begin{align*} \text {integral}& = -\int \left (\frac {1}{2 \sqrt {c+d x}}-\frac {\cosh (2 a+2 b x)}{2 \sqrt {c+d x}}\right ) \, dx \\ & = -\frac {\sqrt {c+d x}}{d}+\frac {1}{2} \int \frac {\cosh (2 a+2 b x)}{\sqrt {c+d x}} \, dx \\ & = -\frac {\sqrt {c+d x}}{d}+\frac {1}{4} \int \frac {e^{-i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx+\frac {1}{4} \int \frac {e^{i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx \\ & = -\frac {\sqrt {c+d x}}{d}+\frac {\text {Subst}\left (\int e^{i \left (2 i a-\frac {2 i b c}{d}\right )-\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{2 d}+\frac {\text {Subst}\left (\int e^{-i \left (2 i a-\frac {2 i b c}{d}\right )+\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{2 d} \\ & = -\frac {\sqrt {c+d x}}{d}+\frac {e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {e^{2 a-\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=-\frac {\sqrt {c+d x}}{d}+\frac {e^{2 a-\frac {2 b c}{d}} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {c+d x}}-\frac {e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {c+d x}} \]

[In]

Integrate[Sinh[a + b*x]^2/Sqrt[c + d*x],x]

[Out]

-(Sqrt[c + d*x]/d) + (E^(2*a - (2*b*c)/d)*Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2, (-2*b*(c + d*x))/d])/(4*Sqrt[2]*
b*Sqrt[c + d*x]) - (E^(-2*a + (2*b*c)/d)*Sqrt[(b*(c + d*x))/d]*Gamma[1/2, (2*b*(c + d*x))/d])/(4*Sqrt[2]*b*Sqr
t[c + d*x])

Maple [F]

\[\int \frac {\sinh \left (b x +a \right )^{2}}{\sqrt {d x +c}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=\frac {\sqrt {2} \sqrt {\pi } {\left (d \cosh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - d \sinh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) - \sqrt {2} \sqrt {\pi } {\left (d \cosh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + d \sinh \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) - 8 \, \sqrt {d x + c} b}{8 \, b d} \]

[In]

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

[Out]

1/8*(sqrt(2)*sqrt(pi)*(d*cosh(-2*(b*c - a*d)/d) - d*sinh(-2*(b*c - a*d)/d))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x + c
)*sqrt(b/d)) - sqrt(2)*sqrt(pi)*(d*cosh(-2*(b*c - a*d)/d) + d*sinh(-2*(b*c - a*d)/d))*sqrt(-b/d)*erf(sqrt(2)*s
qrt(d*x + c)*sqrt(-b/d)) - 8*sqrt(d*x + c)*b)/(b*d)

Sympy [F]

\[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=\int \frac {\sinh ^{2}{\left (a + b x \right )}}{\sqrt {c + d x}}\, dx \]

[In]

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

[Out]

Integral(sinh(a + b*x)**2/sqrt(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.77 \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=\frac {\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (2 \, a - \frac {2 \, b c}{d}\right )}}{\sqrt {-\frac {b}{d}}} + \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-2 \, a + \frac {2 \, b c}{d}\right )}}{\sqrt {\frac {b}{d}}} - 8 \, \sqrt {d x + c}}{8 \, d} \]

[In]

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

[Out]

1/8*(sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(-b/d))*e^(2*a - 2*b*c/d)/sqrt(-b/d) + sqrt(2)*sqrt(pi)*er
f(sqrt(2)*sqrt(d*x + c)*sqrt(b/d))*e^(-2*a + 2*b*c/d)/sqrt(b/d) - 8*sqrt(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=-\frac {{\left (\frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b d} \sqrt {d x + c}}{d}\right ) e^{\left (\frac {2 \, b c}{d}\right )}}{\sqrt {b d}} + \frac {\sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {-b d} \sqrt {d x + c}}{d}\right ) e^{\left (-\frac {2 \, {\left (b c - 2 \, a d\right )}}{d}\right )}}{\sqrt {-b d}} + 8 \, \sqrt {d x + c} e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, a\right )}}{8 \, d} \]

[In]

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

[Out]

-1/8*(sqrt(2)*sqrt(pi)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)/d)*e^(2*b*c/d)/sqrt(b*d) + sqrt(2)*sqrt(pi)*d*er
f(-sqrt(2)*sqrt(-b*d)*sqrt(d*x + c)/d)*e^(-2*(b*c - 2*a*d)/d)/sqrt(-b*d) + 8*sqrt(d*x + c)*e^(2*a))*e^(-2*a)/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^2}{\sqrt {c+d\,x}} \,d x \]

[In]

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

[Out]

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