Optimal. Leaf size=268 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (-\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )} \]
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Rubi [A] time = 0.269515, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {747, 401, 108, 409, 1218, 444, 63, 212, 208, 205} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\left (a^2 d+b^2 c\right )^{3/4}}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (-\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )} \]
Antiderivative was successfully verified.
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Rule 747
Rule 401
Rule 108
Rule 409
Rule 1218
Rule 444
Rule 63
Rule 212
Rule 208
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{(a+b x) \left (c+d x^2\right )^{3/4}} \, dx &=a \int \frac{1}{\left (a^2-b^2 x^2\right ) \left (c+d x^2\right )^{3/4}} \, dx-b \int \frac{x}{\left (a^2-b^2 x^2\right ) \left (c+d x^2\right )^{3/4}} \, dx\\ &=-\left (\frac{1}{2} b \operatorname{Subst}\left (\int \frac{1}{\left (a^2-b^2 x\right ) (c+d x)^{3/4}} \, dx,x,x^2\right )\right )+\frac{\left (a \sqrt{-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-\frac{d x}{c}} \left (a^2-b^2 x\right ) (c+d x)^{3/4}} \, dx,x,x^2\right )}{2 x}\\ &=-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{a^2+\frac{b^2 c}{d}-\frac{b^2 x^4}{d}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{d}-\frac{\left (2 a \sqrt{-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-b^2 c-a^2 d+b^2 x^4\right ) \sqrt{1-\frac{x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{x}\\ &=-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2 c+a^2 d}-b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\sqrt{b^2 c+a^2 d}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2 c+a^2 d}+b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\sqrt{b^2 c+a^2 d}}+\frac{\left (a \sqrt{-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{b x^2}{\sqrt{b^2 c+a^2 d}}\right ) \sqrt{1-\frac{x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\left (b^2 c+a^2 d\right ) x}+\frac{\left (a \sqrt{-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{\sqrt{b^2 c+a^2 d}}\right ) \sqrt{1-\frac{x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{\left (b^2 c+a^2 d\right ) x}\\ &=-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}-\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\left (b^2 c+a^2 d\right )^{3/4}}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (-\frac{b \sqrt{c}}{\sqrt{b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x}+\frac{a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (\frac{b \sqrt{c}}{\sqrt{b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{\left (b^2 c+a^2 d\right ) x}\\ \end{align*}
Mathematica [A] time = 0.201889, size = 237, normalized size = 0.88 \[ -\frac{\sqrt{b} x \sqrt [4]{a^2 d+b^2 c} \left (\tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )+\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )\right )+a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (-\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .-\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )+a \sqrt [4]{c} \sqrt{-\frac{d x^2}{c}} \Pi \left (\frac{b \sqrt{c}}{\sqrt{d a^2+b^2 c}};\left .-\sin ^{-1}\left (\frac{\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{x \left (a^2 d+b^2 c\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.615, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{bx+a} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{2} + c\right )}^{\frac{3}{4}}{\left (b x + a\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right ) \left (c + d x^{2}\right )^{\frac{3}{4}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (d x^{2} + c\right )}^{\frac{3}{4}}{\left (b x + a\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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