Optimal. Leaf size=345 \[ \frac{\sqrt{b} \left (\frac{b x^2}{a}+1\right )^{3/4} (3 a d+4 b c) \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{6 \sqrt{a} c \left (a+b x^2\right )^{3/4} (b c-a d)^2}+\frac{b x (3 a d+4 b c)}{6 a c \left (a+b x^2\right )^{3/4} (b c-a d)^2}-\frac{d x}{2 c \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (b c-a d)}-\frac{\sqrt [4]{a} d \sqrt{-\frac{b x^2}{a}} (9 b c-2 a d) \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (b c-a d)^3}-\frac{\sqrt [4]{a} d \sqrt{-\frac{b x^2}{a}} (9 b c-2 a d) \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (b c-a d)^3} \]
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Rubi [A] time = 0.382358, antiderivative size = 345, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {414, 527, 530, 233, 231, 401, 108, 409, 1218} \[ \frac{b x (3 a d+4 b c)}{6 a c \left (a+b x^2\right )^{3/4} (b c-a d)^2}-\frac{d x}{2 c \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (b c-a d)}+\frac{\sqrt{b} \left (\frac{b x^2}{a}+1\right )^{3/4} (3 a d+4 b c) F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 \sqrt{a} c \left (a+b x^2\right )^{3/4} (b c-a d)^2}-\frac{\sqrt [4]{a} d \sqrt{-\frac{b x^2}{a}} (9 b c-2 a d) \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (b c-a d)^3}-\frac{\sqrt [4]{a} d \sqrt{-\frac{b x^2}{a}} (9 b c-2 a d) \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{a d-b c}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c x (b c-a d)^3} \]
Antiderivative was successfully verified.
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Rule 414
Rule 527
Rule 530
Rule 233
Rule 231
Rule 401
Rule 108
Rule 409
Rule 1218
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )^2} \, dx &=-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}+\frac{\int \frac{2 b c-a d-\frac{5}{2} b d x^2}{\left (a+b x^2\right )^{7/4} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}-\frac{\int \frac{\frac{1}{2} \left (-2 b^2 c^2+12 a b c d-3 a^2 d^2\right )-\frac{1}{4} b d (4 b c+3 a d) x^2}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx}{3 a c (b c-a d)^2}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}-\frac{(d (9 b c-2 a d)) \int \frac{1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx}{4 c (b c-a d)^2}+\frac{(b (4 b c+3 a d)) \int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx}{12 a c (b c-a d)^2}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}-\frac{\left (d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-\frac{b x}{a}} (a+b x)^{3/4} (c+d x)} \, dx,x,x^2\right )}{8 c (b c-a d)^2 x}+\frac{\left (b (4 b c+3 a d) \left (1+\frac{b x^2}{a}\right )^{3/4}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/4}} \, dx}{12 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}+\frac{\sqrt{b} (4 b c+3 a d) \left (1+\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 \sqrt{a} c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}+\frac{\left (d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{a}} \left (-b c+a d-d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{2 c (b c-a d)^2 x}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}+\frac{\sqrt{b} (4 b c+3 a d) \left (1+\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 \sqrt{a} c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{\left (d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{\sqrt{d} x^2}{\sqrt{-b c+a d}}\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c (b c-a d)^3 x}-\frac{\left (d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{\sqrt{d} x^2}{\sqrt{-b c+a d}}\right ) \sqrt{1-\frac{x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c (b c-a d)^3 x}\\ &=\frac{b (4 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )}+\frac{\sqrt{b} (4 b c+3 a d) \left (1+\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{6 \sqrt{a} c (b c-a d)^2 \left (a+b x^2\right )^{3/4}}-\frac{\sqrt [4]{a} d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}} \Pi \left (-\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (b c-a d)^3 x}-\frac{\sqrt [4]{a} d (9 b c-2 a d) \sqrt{-\frac{b x^2}{a}} \Pi \left (\frac{\sqrt{a} \sqrt{d}}{\sqrt{-b c+a d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c (b c-a d)^3 x}\\ \end{align*}
Mathematica [C] time = 0.538072, size = 387, normalized size = 1.12 \[ \frac{x \left (\frac{c \left (36 a c \left (6 a^2 d^2+3 a b d \left (d x^2-4 c\right )+2 b^2 c \left (3 c+2 d x^2\right )\right ) F_1\left (\frac{1}{2};\frac{3}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-6 x^2 \left (3 a^2 d^2+3 a b d^2 x^2+4 b^2 c \left (c+d x^2\right )\right ) \left (4 a d F_1\left (\frac{3}{2};\frac{3}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+3 b c F_1\left (\frac{3}{2};\frac{7}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (6 a c F_1\left (\frac{1}{2};\frac{3}{4},1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-x^2 \left (4 a d F_1\left (\frac{3}{2};\frac{3}{4},2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )+3 b c F_1\left (\frac{3}{2};\frac{7}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )\right )}+b d x^2 \left (\frac{b x^2}{a}+1\right )^{3/4} (3 a d+4 b c) F_1\left (\frac{3}{2};\frac{3}{4},1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )}{36 a c^2 \left (a+b x^2\right )^{3/4} (b c-a d)^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{2}+c \right ) ^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{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 (b x^{2} + a\right )}^{\frac{7}{4}}{\left (d x^{2} + c\right )}^{2}}\,{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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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 (b x^{2} + a\right )}^{\frac{7}{4}}{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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